Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download An investigation into the algebraic structure of our numbers.
Document related concepts
Approximations of π wikipedia , lookup
History of logarithms wikipedia , lookup
Infinitesimal wikipedia , lookup
Location arithmetic wikipedia , lookup
Vincent's theorem wikipedia , lookup
List of first-order theories wikipedia , lookup
Large numbers wikipedia , lookup
Abuse of notation wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Positional notation wikipedia , lookup
Factorization wikipedia , lookup
Collatz conjecture wikipedia , lookup
P-adic number wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Transcript
Copyright: Gary Harris, 2009
Advanced Math for Middle School Teachers I
Integers and fractions: An investigation into the algebraic structure
of our numbers. ©
By
Gary A. Harris
A project of the West Texas Middle School Math Partnership
Funded by NSF MSP grant # 0831420.
The opinions expressed herein are those of the WTMSMP personnel and associates and do not necessarily reflect
those of NSF.
1
Copyright: Gary Harris, 2009
Integers and Fractions: An investigation into the algebraic structure of our numbers
Preface
The adoption of the Common Core State Standards for Mathematics (CCMS) for students
from elementary grades through high school has caused considerable attention to be focused on
the mathematical content preparation of math teachers, especially at the middle school level. The
recently published Conference board of the Mathematical Sciences document titled “The
Mathematical Education of Teachers II” (CBMS MET II) concludes that there is evidence that
prospective middle school math teachers do not receive sufficient coursework in the discipline of
mathematics. The MET II authors argue that the future teachers need courses that allow them to
delve into the mathematics of the middle grades, strengthen their own knowledge, and let them
connect the math taught through elementary, middle, and high school. (We would add “into
college.”) Moreover, they empathically state that no such course should be at or below the level
of pre-calculus. This course text is designed to meet this challenge with regard to providing the
future teachers the opportunity to develop a deep understanding of the algebraic structure of the
real number field.
The MET II writers point out that “a strong foundation in work with rational numbers is
absolutely essential for teaching in the middle grades.” Upon completion of this course material
the future teacher should meet each of the following criteria specifically mentioned in the MET
II document:
Understand and be able to explain the mathematics that underlies the procedures used for
operating on whole numbers and rational numbers.
Understand and be able to explain the distinctions among whole numbers, integers, rational
numbers, and real numbers.
Convert easily among fractions, decimals, and percents.
Demonstrate facility in using number and operation properties.
Apply proportions appropriately and provide explanations.
Understand how decimals extend the place value work from the earlier grades.
Understand why only repeating decimals can be converted to fractions, and why
nonrepeating decimals are not rational.
Demonstrate algebraic skills and be able to give a rationale for common algebraic
procedures.
In addition we add the following:
Understand the importance the geometric series in decimal representations.
Understand the concept of “limiting value” as it pertains to an infinitely repeating
decimal representing a rational number.
Understand the importance of the Least Upper Bound Principle in the existence of
irrational numbers.
Explain every individual step in simplifying a rational expression.
Explain why all the rules for placement of a minus sign are required.
Text in a nut shell. This text is suitable for use in a mathematics course for pre-service middle
school teachers offered as an upper division undergraduate level math course, or as a lower
division graduate level math course. The text begins with the development of the natural
2
Copyright: Gary Harris, 2009
numbers and explores the operation of addition with particular attention paid to the special
properties of 0. The concept of the additive inverse leads to a study of the meaning of negative
numbers and subtraction, culminating in the group of integers. The operations of multiplication,
extended to include negative integers leads to the ring of integers. The inclusion of
multiplicative inverses leads to the study of rational number field, with significant time spent on
equivalence of fractions, and multiplication and addition. The division algorithm and the decimal
representation of rational numbers are covered in depth, including the relevance of the Geometric
Series. The argument that there is no rational number whose square is 2 leads to a study of
irrational numbers, including their existence as a consequence of the Least Upper Bound
Principle.
Text Logical Progression.
Chapter 1. The Abelian Group of Integers;
Definitions and Structure
Chapter 2. The Commutative Ring of Integers with Unity;
Definitions and Structure
Chapter 3. The Rational Number Field; Definition and Structure
Chapter 4. Decimal Representations of Rational numbers
Chapter 5. The Real Number Field; Existence of Irrational Numbers
3
Copyright: Gary Harris, 2009
Table of Contents
Chapter 1: The group of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 6
Chapter 1 begins with the definition of a “positive integer” and
introduces the operation of addition of positive integers, observing that
addition is commutative and associative. Then the additive identity
(“0”) is discussed. The chapter proceeds to provide an axiomatic
development of the additive inverses (“negative integers”) and extends
the operation of addition to them so as to preserve commutativity and
associativity.
Section 1.1: The positive integers . . . . . . . . . . . . . . . . . . . . . .pg 7
Exercises 1.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .pg 9
Section 1.2: Addition of positive integers . . . . . . . . . . . . . . . .pg 10
Exercises 1.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..pg 12
Section 1.3: Negative integers. . . . . . . . . . . . . . . . . . . . . . . . . pg 13
Exercises 1.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..pg 15
Section 1.4: Addition of negative integers. . . . . . . . . . . . . . . .pg 17
Exercises 1.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..pg 18
Section 1.5: Subtraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .pg 19
Exercises 1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..pg 20
Chapter 2: The ring of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 21
Chapter 2 defines multiplication of positive integers as multiple
addition and observes the distributive property of multiplication over
addition. Multiplication is then extended to the negative integers
axiomatically by preserving the commutative, associative, and
distributive properties.
Section 2.1: Multiplication of positive integers. . . . . . . . . . . .pg 22
Exercises 2.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 24
Section 2.2: Multiplication involving negative integers. . . . . pg 26
Exercises 2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 28
Chapter 3: The rational number field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg29
Chapter 3 proceeds to axiomatically define the multiplicative inverses
of positive integers (1/n). Multiplication and addition are extended to
such numbers so as to preserve the commutative, associative, and
distributive properties. All this structure is then extended in similar
fashion to include additive inverses (negative integers and fractions).
Section 3.1: Multiplicative inverse of a positive integer. . . . . pg 30
Exercises 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 32
Section 3.2: Multiplication of positive integer fractions. . . . . pg 33
Exercises 3.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 35
Section 3.3: Addition of positive integer fractions. . . . . . . . .. pg 37
4
Copyright: Gary Harris, 2009
Exercises 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 42
Section 3.4: Fractions with negative integers. . . . . . . . . . . . . .pg 43
Exercises 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 44
Chapter 4: Representations of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 45
The usual equivalence relation of fractions is discussed as well as the
reduced forms of fractions. The Division Algorithm is introduced to
convert fractions to decimal form. It is observed that the resulting
decimal representation must either be finite or exhibit an infinitely
repeating pattern. The geometric series and concept of limiting value
is introduced to derive the meaning of the statement: “A fraction has
an infinite decimal representation”. The formula for the partial sum of
a geometric series is used to “convert” infinite decimals expressions to
the fraction. Much emphasis is placed on the fact that the word “equal”
as used in the expression ½ = 0.5 is not the same as the word “equal”
as used in the expression 1/3 = 0.33333333…..
Section 4.1: Equivalence of fractions. . . . . . . . . . . . . . . . . . . .pg 46
Exercises 4.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 51
Section 4.2: Fractions to decimal form. . . . . . . . . . . . . . . . . . pg 52
Exercises 4.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 57
Section 4.3: Decimal to fraction form. . . . . . . . . . . . . . . . . . . pg 58
Exercises 4.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 63
Chapter 5: Other kinds of numbers?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 64
The usual argument that there is no fraction with square equal to 2 is
extended to a determination of exactly when it is possible for a fraction
to the pth power to equal a given rational number. It is shown that the
existence of the square root of two is a consequence of the assumption
of an axiom known as the Least Upper Bound Principle. Moreover, it
is shown that the assumption of this axiom implies that any infinite
decimal “represents” a number. This leads to the existence of numbers
that cannot be represented as a fraction of integers.
Section 5.1: Square root of 2. . . . . . . . . . . . . . . . . . . . . . . . . . pg 65
Exercises 5.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 68
Section 5.2: The least upper bound principal (irrationals). .. . pg 69
Exercises 5.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 72
Section 5.3: The quadratic formula . . . . . . . . . . . . . . . . . . . . pg 73
Exercises 5.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . pg 74
5
Copyright: Gary Harris, 2009
Chapter 1: The Group of Integers
Introduction.
The purpose of this text is to guide us on an exploration of our number system with
emphasis on the definitions of the numbers, how the numbers are combined to produce other
numbers, and the underlying abstract algebraic structure that governs all these interactions. In
order to understand our number system and its algebraic structure, it is necessary for us to
understand the fundamental construct on which it is founded: the positive integers. Positive
integers are familiar to children from pre-school on, however their meaning is inherently abstract
and considerably more profound than most of us would anticipate. So this is where Chapter 1
begins. The following Chapter 1 concept map is intended to guide our explorations into the
world of the Group of Integers.
Chapter 1.
The Group of Integers
Whole Numbers
Natural Numbers
Positive Integers:
Operation: Addition
Algebraic Structure:
Commutative and Associative
Non-Negative Integers:
Operation: Addition
Algebraic Structure:
Additive Identity
Commutative and Associative
Define “0”
Define
Integers:
Operation: Addition
Algebraic Structure:
Additive Identity
Additive Inverses
Commutative and Associative
6
Copyright: Gary Harris, 2009
Section 1.1.
We begin our exploration with the following seemly simple question: What is a positive
integer?
A first place to look for the answer to this question might be our dictionary. The
dictionary I am using is WEBSTER’S NEW COLLEGIATE DICTIONARY, G. & C. Merriam
Company, Springfield, Massachusetts, 1980. Let’s start by looking up the definition of the
smallest positive integer, namely 1.
(Dictionary Definitions)
One: The number denoting unity. The first in a set or series. A single person or thing.
This begs for the definition of the term “unity”.
Unity: A quality or state of not being multiple: ONENESS.
for which 1 is made to stand in calculation.
A definite amount taken as one or
This seems a bit circular. Let’s try another number that might be luckier.
Seven: The seventh in a set or series. Something having seven units or members.
The second definition is very curious since it includes the word “seven” which is what we are
looking for in the first place. The first definition begs for the definition of the term “seventh”.
Seventh: A musical interval embracing seven diatonic degrees.
This doesn’t seem to be getting us anywhere. Let’s expand and try to find the definition of
“integer”.
Integer: Any of the natural numbers, the negative of these numbers, or zero.
Ok, this begs for the definition of the term “natural number”.
Natural Number: The number 1, or any number obtained by adding 1 to itself one or more
times: a positive integer.
We seem to be back to square one, whatever “one” might mean. So the next place to look might
be in a mathematics specific dictionary. The one I am using is MATHEMATICS
DICTIONARY: MULTILINGUAL EDITION by James and James, D. Van Nostrand Company,
Princeton, NJ, 1959.
(Math Dictionary Definitions)
One: The Cardinal number denoting a single unit.
7
Copyright: Gary Harris, 2009
So we try again. What is the meaning of “unit”?
Unit: A standard of measurement such as an inch, a foot, a centimeter, a pound, or; a dollar; a
single one of a number, used as the basis of counting or calculating.
This seems a little more physical in nature but still seems circular. After all what does the phrase
“a single one of a number” mean if we don’t know what “one” means?
Let’s try the phrase “Cardinal number”.
Cardinal number: A number which designates the manyness of a set of things; the number of
units, but not the order in which they are arranged; used in distinction to signed numbers. Two
sets are said to have the same cardinal number if their elements can be put into one-to-one
correspondence with each other. Thus a symbol or cardinal number can be associated with any
set. The cardinal number of the set a1 , a2 , a3 , , an is denoted by n.
It is starting to look like we might be forced to accept a level of ambiguity in the
definition of the numbers. However, the phrase “manyness of a set of things” seems to have
some meaning. The “manyness” of the dollars in my bank account means something to me, and
the way this “manyness” changes certainly has meaning to me. Let’s try one more.
Integer: Peano defined the positive integers as a set of elements which satisfies the following
postulates: (1) There is a positive integer 1; (2) every positive integer a has a consequent a+ ( a is
called the antecedent of a+); (3) the integer 1 has no antecedent; (4) if a+ = b+ then a = b; (5)
every set of positive integers which contains 1 and the consequent of every number of the set
contains all the positive integers.
1 1
1 1 1
This version of Peano’s definition is rather strange in that the set S 1, , , , , ,
,
n n 1
2 3 4
1
1
seems to satisfy all the stated conditions. Note that in this set the consequent of is
.
n n 1
This is not at all helpful! Whatever we think about the positive integers, we don’t think they are
the set S.
A second definition appearing in the math dictionary is the following:
Integer: A positive integer (or zero) can also be thought of as describing the “manyness” of a set
of objects in the sense of being a symbol denoting that property of a set of individuals which is
independent of the natures of the individuals. That is, it is a symbol associated with a set and
with all other sets which can be put into one-to-one correspondence with this set. The set is
called a “number class”.
It looks like we are simply going to have to accept the circular nature of the definitions of
the numbers and think of a positive integer n as representing that common property shared by all
collections that contain exactly n objects. For example, the number three (tres, trois, drei), 3,
describes the common property of the collections a, d , y , , , , , , and any other
8
Copyright: Gary Harris, 2009
collection whose objects can be paired (matched up) in a one-to-one fashion with the objects in
any one of these collections. Any of these collections would be called a number class for the
number 3. In other words, the number 3 describes a collection of three gum drops, or a collection
of 3 pencils, or a collection of 3 cars in the parking lot, or a collection containing one gumdrop,
one car, and one pencil. So we just have to accept the fact that the number 3, by itself, is
inherently an abstract concept.
Our ultimate goal is to obtain a deep understanding of the algebraic structure of our
number system. So we want to be able to combine numbers and solve for the value of numbers
given conditions that numbers must satisfy. To begin we need to know how to combine
numbers. We will consider this in Section 1.2, but first let’s do some exercises.
Exercises for Section 1.1.
Exercise 1.1.1. Provide four different number classes, each of which is represented by the
number 3. Label the number classes C1, C2, C3, and C4.
Exercise 1.1.2. What is the number represented by the number class C1 , C2 , C3 , C4 ?
Explain.
Exercise 1.1.3. Which of the following number classes represent the number 2?
1, 2 , , # ,2 ,1, 2 , , #, b, c, d , z, 4 ,{9, Tennessee}
Explain.
(By the term “model” we mean an activity or demonstration of a concept that would be suitable
for use in a middle school math class.)
Exercise 1.1.4. Provide two different models for each of the following integers: 1, 5, 15, 25.
9
Copyright: Gary Harris, 2009
Section 1.2.
Binary operation on a set of numbers: A method for putting two numbers in the set together and
producing a number in the set. So a binary operation on the set of positive integers is a process
for putting two positive integers together to get a positive integer.
Our first binary operation on the set of positive integers is addition.
Addition of Positive Integers.
The sum of two positive integers: Let m and n be two positive integers. Pick a number class for
the integer m (a set of exactly m elements) and a number class for the integer n (another set of
exactly n elements). We define the sum, m + n, to be the positive integer represented by the
number class gotten by combining the number class of m with the number class of n. It is
implicit in this definition that two number classes do not overlap, even if they are number classes
for the same integer.
For example we can choose the number class consisting of a cup containing exactly 5
jellybeans to represent the integer 5 and the number class consisting of another cup containing
exactly 3 jellybeans to represent the integer 3. We pour the two cups of jellybeans into a third
cup and let 5 + 3 be the integer represented by the number class consisting of the third cup
containing exactly 8 jellybeans. We don’t count the elements in the number class for 5 + 3 until
after the elements from the number class for 5 and the elements from the number class for 3 are
combined into the third cup. So it doesn’t matter which cup, the one for 5 or the one for 3, got
poured into the third cup first. Of course this fact has nothing to do with the particular numbers 5
and 3, nor with the type of objects being counted. We could combine the set of 5 jellybeans with
a set of 3 thumb tacks and get a set of 8 objects. However, in practice this might not be a good
idea.
Fact 1.2.1. Addition of positive integers is commutative, meaning if m and n are two positive
integers then m + n = n + m.
Suppose we have three positive integers m, n, and p and we want the result of adding all
three. Since addition is defined as a binary operation, a way to combine two elements at a time,
we have to decide which two to add first and then add the resulting integer to the third. For
example we could add m and n to get m + n and then add the resulting integer to p. The notation
for this is m n p . Or we could add n and p first, and then add the resulting integer to m.
The notation for this statement would be n p m . Since addition is commutative, this last
expression is the same as m n p .
For example, we can combine the cup of jellybeans represented by 5 and the cup
represented by 3 to get the cup represented by 8 , (We have only two hands, so we can pick up
only two cups at a time and pour them together.) and then combine it with the cup represented by
4 to get the cup represented by 5 3 4 , which is represented by the integer 12. But we could
have first combined the cups represented by 3 and 4 respectively to get a number class for 7 (a
cup of 7 jellybeans) and then combine that with the cup of 5 to get a number class for 5 3 4
10
Copyright: Gary Harris, 2009
which also represents the integer 12. As above, this has nothing to do with the particular positive
integers.
Fact 1.2.2. Addition of positive integers is associative, meaning if m, n, and p are three positive
integers then m n p = m n p .
Any time we write an expression like 5 + 2 + 3 or m + q + r (with no parentheses) we are
using associativity because it does not matter which pair we add first.
The Number 0.
We define a new integer zero, 0, to represent a number class that has nothing in it. For
example, the set of all cows that can flap their ears and fly is a number class for the integer 0.
Also an empty checking account could represent a number class for 0. Note that it does not
matter what we call the elements in a number class for 0 since there is nothing in the set. Note
that if we combine a set with no jellybeans with a set that has 15 jellybeans we get a set with 15
jellybeans. In general, if we combine a set with n elements with a set that contains no elements,
we get a set that contains n elements.
Fact 1.2.3. Zero is an additive identity for the positive integers, meaning if m is a positive
integer then m + 0 = m. (Commutativity also gives us 0 + m = m.)
We defined 0 in terms of its number class and defined addition in terms of combining
number classes. It is obvious that
, an empty set combined with an empty set is still an
empty set. So 0 is an additive identity on our set of numbers (positive integers and 0). We will
require that 0 is an additive identity on any numbers we wish to add to our set of numbers. (See
Section 1.3 for an example of this requirement.) It is conceivable that there might be some other
abstract number which is also an additive identity on our set of numbers. But, in fact, this is not
the case.
Fact 1.2.4. There can be only one additive identity.
Proof: Suppose that 01 is an additive identity for our set of numbers. Then,
01+ 0 = 0
since 01 is an additive identity. Also
01+ 0 = 01
since 0 is an additive identity. So
01 = 0.
Perhaps a nicer way to present this would be the following:
0 = 01+ 0
since 01 is an additive identity, and
01+ 0 = 01
since 0 is an additive identity. So
0 = 01 .
Here is our first use of the word “proof”.
11
Copyright: Gary Harris, 2009
Definition of “Proof”. A sequence of logically valid steps that uses definitions, facts, or agreed
upon assumptions to obtain a particular conclusion, which will then be considered as a fact.
In the case of Fact 1.2.4 we have used the definition of additive identity to conclude that there is
only one additive identity for our set of numbers. Henceforth we will refer to 0 as the additive
identity for our set of numbers.
Exercises for Section 1.2.
Read the articles by Russell and Conant (Supplied with text.).
Recall from the exercise for Section 1.1 that we use the term “model” to refer to an activity or
demonstration of a concept that would be suitable for use in a middle school math class.
Exercise 1.2.1. Provide two different models for addition of two positive integers.
Exercise 1.2.2. Provide two different models that demonstrate that addition of positive integers
is commutative and associative.
Exercise 1.2.3. Provide two different models for the integer zero along with demonstrations that
zero is the additive identity in each case.
12
Copyright: Gary Harris, 2009
Section 1.3.
Negative Integers.
Suppose for the moment that the integer 1 represents the length of our step. So the
number class for 1 might be the set containing a straight stick with length equal the length of our
step. Combining two such number classes would give a number class for 1 + 1 containing two
such sticks which, if laid end to end, would be the distance we travel in 2 steps. Combining three
such number classes would produce a set with three sticks which laid end to end would be the
distance we travel in 3 steps, and so on. Suppose we take m steps and then decide we need to
return to where we started. We would like to turn around and go back the same distance, in
effect canceling the distance we had traveled. This is our motivation for the following abstract
construct:
Definition of additive inverse of a positive integer: For a positive integer m, we define the
additive inverse of m to be the number that added (whatever “added” and “the number” might
mean in this case) to m returns 0.
We use the notation m to denote the additive inverse of the positive integer m . So
n m means n m 0 and m n 0.
We have both equations because we want to make sure that our operation of summation remains
commutative.
Definition of negative integer: A negative integer is a number that is the additive inverse of a
positive integer.
So a is a negative integer provided there is a positive integer b so that a b . In other words, a
is the additive inverse of the positive integer b. This also means that a b 0 for some positive
integer b .
Definition of additive inverse of a negative integer: If n is a negative integer we define the
additive inverse of n to be the integer that added to n returns 0 .
Notice that a b 0 means that a is the additive inverse of b ( a b ) and that b is the additive
inverse of a ( b a ).
Notice that it follows from the above definition that m is the additive inverse of
–m since m is the number that added to –m returns 0, that is (–m) + m = 0. This is really the
“proof” of the well-known statement:
Fact 1.3.1. m (m) .
Proof: m (m) 0.
This proof simply uses the definition of additive inverse; namely, m is the additive inverse of
m simply because m is what you add to m to get 0 .
13
Copyright: Gary Harris, 2009
Keep in mind that –(something) is the symbol for the additive inverse of the something.
We want to extend what we know about addition of positive integers to negative integers. The
properties of commutativity and associativity of addition of positive integers are very useful, so
we make the following
Basic assumption: Our extension of addition to include negative integers must also be
commutative and associative.
We will see that a lot of the well -known rules of arithmetic are results of this requirement.
Fact 1.3.2. 0 is the additive identity for negative integers.
Proof. A negative integer is the additive inverse of some positive integer. Suppose m is a
positive integer. We want to show that 0 (m) m . In other words we want to show that
0 (m) is the additive inverse of m. So we simply add it to m and see if we get 0. We see that
(0 + (–m)) + m = 0 + ((–m) + m)
by associativity. It follows that
(0 + (–m)) + m = 0 + 0
by definition of additive inverse of m. So
(0 + (–m)) + m = 0
since 0 is the additive identity. Thus
(0 + (–m)) =
We have been using the phrase “the additive inverse of m”. This suggests there cannot be
two different additive inverses for the same integer m.
Fact 1.3.3. If m is an integer, then there is only one additive inverse of m.
Proof. Suppose that a and b are both additive inverses of the integer m. So
m a 0 and m b 0 .
Then
ma mb.
So
(m) (m a) (m) (m b)
by the principle of “equal actions”: equal actions on equal objects produce equal results.
In this case our equal action was to add m to each of two equal objects. By associativity
applied to both sides of this equation, we have
((m) m) a ((m) m) b .
By the definition of additive inverse, we have
0a 0b.
14
Copyright: Gary Harris, 2009
Since 0 is the additive identity, we have
a b.
The proof of these two facts is our first use of what we will be calling the “algebraic
structure of the numbers”. We used the principle of equal actions, associativity, and the
existence of the additive identity and additive inverses.
Our next task is to extend the definition of addition of positive integers to include the
addition of negative integers. But first we need to discuss the relative size of two positive
integers.
Ordering of the Positive Integers.
Suppose m and n are two positive integers and let M and N be number classes for m and
n, respectively. So M contains exactly m elements and N contains exactly n elements. We will
say that n is less than m, denoted n m , provided our attempt to make a one-to-one
correspondence between M and N exhausts all the elements of N, but unpaired elements still
remain in M. In other words, there are more elements in M than there are in N.
That 3 < 7 is the point of the next demonstration. It will be useful to keep this
demonstration in mind when we look at the definitions of 7 (3) and 3 (7) in
Section 1.4.
Demonstration 1.3.4.
M = { a, b, c, d, e, f, g }
N = { #, *, & }
So M is a number class for the integer 7 and N is a number class for
the integer 3. We see that our attempt to create a one-to-one
correspondence between M and N leaves four objects in M after
each object in N has been paired with a distinct object in M. So we
have
3 < 7.
Exercises for Section 1.3.
Recall from the exercises for Sections 1.1 and 1.2 that we use the term “model” to refer to an
activity or demonstration of a concept that would be suitable for use in a middle school math
class.
Exercise 1.3.1. Provide two different models of negative integers.
Exercise 1.3.2. Provide two different models for the integer –3.
15
Copyright: Gary Harris, 2009
Exercise 1.3.3. Use each of your models for the integer –3 to show that –3 + 3 = 0.
Exercise 1.3.4. Provide two different models for two integers m and n for which n < m.
Exercise l.3.5. Explain in words why –(–m) = m .
16
Copyright: Gary Harris, 2009
Section 1.4.
Addition Involving Negative Integers.
We want to define addition of a positive integer and a negative integer, i.e define
m (n) for positive integers m and n. Throughout we will insist that our definitions must
preserve the properties of commutativity and associativity of addition. We will do this in three
stages.
In case m = n: –n is the additive inverse of m, so m (n) 0 and (n) m 0.
In case n m : define m (n) to be the positive integer representing the set of unpaired
elements in the number class M after the one-to-correspondence has exhausted number class N.
For example if M is a set of 15 jellybeans and N is a set of 6 jellybeans then m (n) 9 , the
number representing the set of 9 jellybeans left after 6 from M have been paired with the 6 in the
set N. The pairing operation is commutative; it doesn’t matter if we pick the first elements to be
paired from M or from N. So we also have (n) m 9
In case m n : define m (n) to be the additive inverse of n (m) , that is
m (n) (n (m)) .
So if m = 6 and n =15, then m (n) 6 (15) (15 (6)) 9.
Notice that this definition is forced by our insistence that the commutative and
associative properties hold for addition. Namely, if these two properties hold then
m (n) (n (m)) (m (m)) (n (n)) 0 0 0 .
Thus m (n) must be the additive inverse of n (m)) .
Next we must define the addition of two negative integers. To this end, let m and n be
two positive integers. We define
(m) (n) (m n) .
Again this is forced by our insistence that the commutative and associative laws hold for
addition. Namely,
((m) (n)) (m n) ((m) m) ((n) n) 0 0 0 .
So (m) (n) must be the additive inverse of (m n) .
This completes our construction of our first “number system”, the integers with addition.
The integers with addition form a “Group”.
This means the set of integers is a set of objects with a binary operation (+ in this case) that
satisfies the following conditions:
1. The operation is associative.
2. There is an identity element for the operation.
3. Every element has an inverse element (with respect to the operation).
17
Copyright: Gary Harris, 2009
Suppose that we have three integers m, n, and p (positive or negative) and suppose we know
that
mn m p .
Again we will invoke the “equal action principle” which says that equal actions on equal objects
produce equal results. So if we do the same thing to the equal objects m n and m p we will
get equal objects. The thing we want to do to both these equal objects is to add the additive
inverse of m on the left. The equal action principle tells us that we get equal results,
(m) (m n) (m) (m p).
Applying the associative property of addition gives us
((m) m) n ((m) m) p).
Applying the definition of additive inverse of m gives us
0 n 0 p.
Applying the definition of additive identity gives us
n p.
We have just used the algebraic structure of the group of integers to derive the
Cancellation Law for the Group of Integers: If m, n, and p are integers and
m n m p then n p .
Exercises for Section 1.4.
In the following exercises assume the m and n are positive integers.
Exercise 1.4.1. Provide two models, each with different positive integers m and n, for
m (n) in the case when n < m.
Exercise 1.4.2. Provide two models, each with different positive integers m and n, for
m (n) in the case when n > m.
Exercise 1.4.3. Consider the set
following table:
{
} Define the operation + on
+
a
b
c
d
a
a
b
c
d
b
b
c
d
e
c
c
d
e
a
d
d
e
a
b
e
e
a
b
c
Find the additive identity and the additive inverse of each member of
Show that addition is commutative on
18
by the
e
e
a
b
c
d
.
Copyright: Gary Harris, 2009
Section 1.5.
Subtraction.
Now let’s consider the problem of finding an integer which must be added to a given
integer to produce another given integer. For example what must we add to 3 to get 5? We
know the answer to this is 2, for 3 + 2 = 5. In general, suppose m and n are given integers and
we wish to find the integer x that added to m gives n. That is, we want to find the integer x so
that
m x n.
The equal action principle tells us that
(m) (m x) (m) n.
Applying the associative property of addition gives us
((m) m) x (m) n.
Applying the definition of additive inverse of m gives us
0 x (m) n.
Applying the definition of additive identity gives us
x (m) n.
It is interesting to note that our math dictionary gives the following definition:
Subtraction: The process of finding a quantity which when added to one of two given quantities
will give the other.
So the expression x (m) n which, by commutativity of addition is the same as the expression
x n (m) , is the definition of the “subtraction of m from n”. This is often called “n minus
m”, and is usually denoted by n—m . Note that subtraction is a binary operation which is not
commutative, for example
5—3 = 5 (3) 2
where as
3—5 = 3 (5) 2.
We will always be careful to distinguish between the binary operation of subtraction and the
operation of negation, which turns a number into its additive inverse.
To get a well-known identity lets compute the following expression for integers m and n:
n—m +(m—n).
By the definition of subtraction this is the same as
(n (m)) (m (n)) .
Applying associativity of addition we get ((n (m)) m) (n) .
Applying associativity again we get (n ((m) m)) (n) .
By definition of additive inverse we get (n 0) (n) .
By definition of additive identity we get n (n) .
By definition of additive inverse we get 0.
19
Copyright: Gary Harris, 2009
In other words, n—m is the additive inverse of m —n. Using our notation for the additive inverse
we get the well-known expression
Fact 1.5.1. (n—m)= –(m—n).
In this chapter we have just seen several very typical examples of using the algebraic
structure of a group to derive general formulas, as well as to solve for the value of numbers given
conditions that numbers must satisfy. And we have seen that using the group structure of the
integers we can always solve for the value of an integer x so that
x+m=n
for any values of the integers m and n. To solve more complicated equations we will need more
algebraic structure.
Exercises for Section 1.5.
All the following models and demonstrations should be suitable for use in an elementary/middle
school classroom
Exercise 1.5.1. Provide two models and demonstrations for the meaning of
5—2 = 5 + (–2)
Exercise 1.5.2. Provide two models and demonstrations for the meaning of
2—5 = 2 + (–5)
Exercise 1.5.3. Provide two models and demonstrations for showing that
5—2 = (2—5)
Exercise 1.5.4. Experiment with the TI-83 calculator to see if it distinguishes between the
binary operation of subtraction and the operation of negation.
Exercise 1.5.5. Solve for the value of x in the following equations. Show all your steps,
indicating which properties and definitions you used at each step.
i. 4 x 3
ii. x 6 2
iii.
(Notice that the minus sign in front of x does not have the same meaning as
the minus sign in front of 6. This could be a point of confusion for students.)
Exercise 1.5.6. Johnny has performed the following calculation:
When asked to
explain how he arrived at this answer, Johnny said that he had 3 red chips that represent -3 and
he took two of them away leaving 1 red chip that represents -1. How do you respond?
20
Copyright: Gary Harris, 2009
Chapter 2: The Ring of Integers
Introduction.
In Chapter 1 we constructed the Group of Integers with the operation of addition. We
now want to define a second operation on the integers, multiplication. We will begin by
considering the algebraic structure resulting from the inclusion of multiplication on the positive
integers. Multiplication will then be extended to the negative integers so as to preserve all the
underlying structure. The following Chapter 2 concept map provides an outline for the
progression of the development of the Ring of Integers.
Chapter 2.
The Ring of Integers
Whole Numbers with Multiplication
Whole Numbers
Non-Negative Integers:
Operation: Addition,+
Algebraic Structure:
Additive Identity
Commutative
Associative
Define
Multiplication,
Non-Negative Integers:
Operations:
Addition & Multiplication
Algebraic Structure:
Additive Identity
Multiplicative Identity
Associative (+ and )
Commutative (+ and )
Distributive ( over +)
Extend to Negative Integers
Integers:
Operation: Addition & Multiplication
Algebraic Structure:
Additive Identity & Multiplicative Identity
Additive Inverses
Addition Commutative
Addition Associative
Multiplication Commutative
Multiplication Associative
Multiplication Distributive over Addition
21
Copyright: Gary Harris, 2009
Section 2-1.
Now we give our second binary operation for positive integers, multiplication.
Multiplication of Positive Integers
Definition of the product of two positive integers: Let m and n be two positive integers. The
product of m and n will be denoted by m n and is defined by
m times
This definition, no matter how simplistic it seems at first sight, requires quite a bit of
discussion. For one thing, addition is a binary operation (we add two integers at a time) and we
have not indicated which pair of n’s gets added first, then second, etc. But recall that addition is
associative, so it does not matter how we pair the terms. This is why we can leave off the
parentheses in this expression.
If follows immediately that if m is a positive integer, then 1 m m and m 1 m.
Fact 2.1.1. One is the multiplicative identity for the positive integers, meaning for any positive
integer m, 1 m m m 1.
Also notice that if we combine 100 sets, each of which have no jellybeans in them then
we get a set with no jellybeans in it. So 100 0 0. We will define 0 100 to be 0 to maintain
commutativity, and besides adding a set to itself no times should not produce a set with anything
in it.
Fact 2.1.2. Zero times any positive integer is 0.
Suppose m and n are positive integers (neither are 0). Then m 1 and n has at least one
object in a number class for it. So m n n n n n (m times) must have at least one
object in its number class.
Fact 2.1.3. If m 0 and n 0 then m n 0 . Equivalently, if m n 0 , then either m 0 or
n 0 (or both).
From this definition of multiplication, as presented, it is not immediately obvious that
multiplication of positive integers is commutative. So lets think of a particular way to keep track
of what it means to add n to itself m times. We know n represents any set containing n objects.
Suppose our set is a set containing n squares each with side length 1. So m n is the number of
squares when m such sets of n squares are combined. First lay one set of n squares in a row (side
by side touching each other), then place the second set of n squares in a row lying beside the first
row of n squares. Then place the third row of n squares beside the second row. Keep doing this
until all m sets have been used as rows of n squares. The resulting figure is an m by n rectangle
made up of n m squares. Now consider n m . Let the set represented by m be a set containing
m squares with side length 1. The integer n m represents the set gotten by combining n sets,
22
Copyright: Gary Harris, 2009
each containing m squares of side length 1. The same process as above, using n rows each
containing m squares, creates an n by m rectangle made up of n m squares. But these two
rectangles are made up of the same number of squares.
Fact 2.1.4. Multiplication of positive integers is commutative, meaning if m and n are two
positive integers then m n n m
Suppose m, n, and p are positive integers. Now suppose the number classes are all sets of
cubes with side length 1. Then stacking the cubes as we stacked the squares above, we see that
n p is the number of cubes in the box with n by p rectangular face and depth 1. Now consider
m (n p) . This number is gotten by combining the n by p by 1 rectangular boxes m times and
counting all the resulting cubes. We count these by placing the m rectangular boxes on top of
each other. This produces a rectangular box containing m (n p) cubes. Similarly (m n) p
is the number of cubes in a rectangular box consisting of p rectangular boxes each containing
m n cubes and stacked one on top of the other. The two boxes contain the same number of
cubes since they have the same dimensions.
Fact 2.1.5. Multiplication of positive integers is associative, meaning if m, n, and p are three
positive integers then m (n p) (m n) p
Suppose m, n, and p are positive integers. Consider the expression
m (n q) (n q) (n q) (n q)
m times
Since addition is associative we can ignore the parentheses, and since addition is commutative
we can write this expression as
m (n q) (n n n) (q q q) m n m q
m times
m times
23
Copyright: Gary Harris, 2009
Fact 2.1.6. Multiplication of positive integers distributes over addition of positive integers,
meaning if m, n, and q are three positive integers, then
m (n q) (m n) (m q) .
That multiplication distributes over addition can be demonstrated by using squares of side
length 1, much as was commutativity. For example
+
Summary of Key Facts for Positive Integers.
Multiplication is commutative.
Multiplication is associative.
Multiplication distributes over addition.
1 is the multiplicative identity.
0 times anything positive integer is 0.
Exercises for Section 2.1.
Exercise 2.1.1. Provide two different models for the multiplication of two positive integers.
Exercise 2.1.2. Use your models to demonstrate that multiplication of positive integers is
commutative.
Exercise 2.1.3. Use your models to demonstrate that multiplication is associative.
Exercise 2.1.4. Use your models to demonstrate that multiplication distributes over addition.
Exercise 2.1.5. Provide a verbal explanation for why 0 m 0 for any positive integer m.
Exercise 2.1.6. Provide a verbal explanation for why m n 0
for two positive integers m and n.
24
Copyright: Gary Harris, 2009
Exercise 2.1.7. Use the students in this class to demonstrate that
might consider the following arrays of students.)
3 number classes for 5
:
5 number classes for 3
25
(Hint. You
Copyright: Gary Harris, 2009
Section 2-2.
Multiplication Involving Negative Integers.
We now proceed to define multiplication involving negative integers. The above
definition for positive integers causes problems. For suppose m and n are positive integers and
we want to define (m) n . We don’t know what it would mean to “add n to itself m times”.
But we do know what it means for multiplication to distribute over addition and this is a property
we want to enforce. In all that follows we will require that commutativity, associativity, and
the distributive property hold and then see what formulas and “rules” must come about.
Observe that if multiplication distributes over addition then we must have
((m) n) (m n) ((m) m) n 0 n 0 .
This means that (m) n is the additive inverse of m n . So we have the rule
(m) n (m n) .
For similar reasons we have the rule
m (n) (m n) .
Fact 2.2.1. For positive integers m and n, we have (m) n (m n) m (n) .
Since m and n are positive integers we know that m n is a positive integer. So its
additive inverse, (m n), is a negative integer. We also know that m n n m , so it follows
that (m) n n (m). So this multiplication is commutative.
Fact 2.2.2. A positive integer times and negative integer is a negative integer.
Again let m and n be positive integers. If multiplication distributes over addition we must
have
((m) (n)) (m (n)) ((m) m) (n) 0 (n) (0 n) 0 0 .
This means that (m) (n) is the additive inverse of (m (n)) , so
(m) (n) (m (n))
But from Fact 2.2.1 we have
m (n) (m n) .
The Principle of Substitution: If two integers are equal, either can be substituted into an
expression for the other without changing the validity of the expression.
By the principle of substitution, we have that
(m) (n) ((m n)) .
We have already seen in Fact 1.3.1 that the additive inverse of the additive inverse of an integer
is the original integer. So
26
Copyright: Gary Harris, 2009
(m) (n) m n
Fact 2.2.3. A negative integer times a negative integer is a positive integer.
By commutativity of multiplication of positive integers we see that
(m) (n) m n n m (n) (m) .
Fact 2.2.4. Multiplication of the integers (including negative integers) is commutative.
Suppose m, n, and p are positive integers. Consider the following:
((m) n) p
((m n) p)
((m n) p)
(m (n p)) .
(m) (n p) .
So we have just demonstrated that multiplication involving one negative and two positive
integers is associative. In a similar way we could demonstrate that multiplication involving two
negative and one positive is associative, as well as multiplication of three negative integers is
associative. We see that everything we have is consistent with our requirement that
multiplication be associative.
The integers with addition, +, and multiplication, form a commutative ring with unit.
That is to say, the integers are a set with two binary operations, addition and multiplication. The
set is a commutative group with respect the operation +. The operation is commutative and
associative and distributes over +. Also there is a multiplicative unity (multiplicative identity),
namely the integer 1.
Recall from Fact 2.1.3 that
if m > 0 and n > 0 then m n 0 .
All the multiplications involving one or two negative integers really boil down to computing the
multiplication with the two positive integers and then choosing the appropriate additive inverse.
The following fact follows:
Fact 2.2.5. If m and n are integers (positive or negative) and m n 0 then either m 0 or n 0
(or both).
This lets us solve a fundamentally different type of equation than we could just with the
group structure on the integers. Consider the equation ( x m) ( x n) 0
for integers m and n. Fact 2.2.5 tells us that either the integer x m 0 in which case x m , or
x n 0 in which case x n . So we get two solutions to this equation.
However, within the ring of integers (with all the algebraic structure it has) we cannot find a
solution to the equation (2 m) 3 6 . To be able to deal with equations like this we need to
extend our concept of numbers beyond just the integers.
27
Copyright: Gary Harris, 2009
Exercises for Section 2-2.
All the following models and demonstrations should be suitable for use in an elementary/middle
school classroom.
Exercise 2.2.1. Provide two models that demonstrate the meaning of a negative integer times a
positive integer.
Exercise 2.2.2. Provide two models that demonstrate the meaning of a negative integer times a
negative integer.
Exercise 2.2.3. Show that 1 is the multiplicative identity for the negative integers. Show all the
steps, giving the property or definition used for each step.
Exercise 2.2.4. Show that 0 (6) 0 . Show all the steps, giving the property or definition used
for each step.
Exercise 2.2.5. Verify that ( x 2) ( x 1) x x x 2 . Show all the steps, giving the property
or definition used for each step.
Exercise 2.2.6. Solve for all the values of x that satisfy the equation x x x 2 0 . Show all
the steps, giving the property or definition used for each step. Also explain how you know that
you have found all the possible values of x that satisfy the equation.
Exercise 2.2.7. Show that m (n) (m n) for any integers m and n. Show all the steps,
giving the property or definition used for each step.
Exercise 2.2.8. Show that (m) ((n) p) ((m) (n)) p follows from associativity of
multiplication for positive integers m , n, p. Show all the steps, giving the property or definition
used for each step.
28
Copyright: Gary Harris, 2009
Chapter 3: The Rational Number Field
Introduction.
In Chapter 2 we constructed the Ring of Integers with the operations of addition and
multiplication, along with its underlying algebraic structure. We observed that even with all this
structure we still cannot solve the simple equation
. In order to solve such equations
we need one more bit of algebraic structure: the concept of multiplicative inverses. In Chapter 3
we define the multiplicative inverse of a positive integer and extend the operations of addition
and multiplication to include multiplicative inverses of positive integers so as to preserve the
existing algebraic structure. We then extend the concept of multiplicative inverse to the full Ring
of Integers, including the negative integers, thus completing the Rational Number Field. We
outline this progression in the Chapter 3 concept map.
Chapter 3.
The Rational Number Field
Positive Integer Fractions
Natural Numbers with Multiplication
Positive Integers:
Operations:
Addition & Multiplication
Algebraic Structure:
Additive Identity
Multiplicative Identity
Associative (+ and )
Commutative (+ and )
Distributive ( over +)
Define
Multiplicative
Inverse
Fractions:
Operations:
Addition & Multiplication
Algebraic Structure:
Additive Identity
Multiplicative inverse
Multiplicative Identity
Associative (+ and )
Commutative (+ and )
Distributive ( over +)
Extend Multiplicative Inverse
To all Non-Zero Integers
Fractions:
Operation: Addition & Multiplication
Algebraic Structure:
Additive Identity & Multiplicative Identity
Additive Inverses
Multiplicative Inverses
Addition Commutative
Addition Associative
Multiplication Commutative
Multiplication Associative
Multiplication Distributive over Addition
29
Copyright: Gary Harris, 2009
Section 3.1. Multiplicative inverse of a positive integer.
Our motivation is to extend the ring of integers and create a number system with binary
operations addition, +, and multiplication, , which are both associative and commutative with
multiplication distributing over addition, in which we can solve equations such as (2 x) 3 8
for x. We know that 0 is the only number that acts as the additive identity for the integers and 1
is the only number that acts as the multiplicative identity for the integers. So they are the only
candidates for the respective identity elements for any extended ring. Suppose there is a number
b so that
b 2 1 .
Then by the Equal Action Principle applied to
(2 x) 3 8
we would have
b ((2 x) 3) b 8 .
Next we have
(b (2 x)) (b 3) b 8 .
by the distributive property. Next we can use associativity to get
((b 2) x) (b 3) b 8 .
By our assumption that b 2 1 we have
(1 x) (b 3) b 8 .
Since 1 is the multiplicative identity we have
x (b 3) b 8 .
This looks like the kind of equation we can solve using the group properties of the integers,
except that b is not an integer, so b 3 may not be an integer. All we would have to do is add the
additive inverse of b 3 , what every that might be in this case, to both sides of the equation and
get
x (b 8) ((b 3)) .
So we have “solved for x”.
The key to solving equations like the one above is the existence of the magic number b
that multiplies times an integer m to get 1 and, of course, that 1 is the multiplicative identity for
the extended ring, 0 is the additive identity, and multiplication and addition are both associative.
In all that follows we will require that multiplication and addition be associative and
commutative, and that multiplication distributes over addition.
Definition of multiplicative inverse: If m is a positive integer, we define the multiplicative
inverse of m to be the number a with the property that a m 1 m a .
We will use the notation
1
to represent the multiplicative inverse of m. We immediately obtain
m
the following
Fact 3.1.1. m
1
1
1 m.
m
m
30
Copyright: Gary Harris, 2009
Of course this definition is curious since we have not yet clearly defined what we mean
by times, , in this context. There is a geometric construction known to the early Greeks that
1
gives us a nice geometric representation for the “number”
as well as the possible meaning of
m
the operation . In particular, given any positive integer m, the Greeks knew how to use
compass and straight edge to divide a line segment of any given length into m line segments of
equal length. For example suppose L is a line segment of “length” 1 unit (foot, mile, kilometer,
whatever). Suppose we wish to divide L into 3 segments of equal length. Let A denote the left
end point of L and let B denote the right end point of L. Draw a line beginning at point A and
making some positive angle with L. With your compass construct 3 adjoining line segments of
equal length beginning at A and lying along this line. Label the endpoint of the last of these
segments C. Use your straight edge to draw the line segment connecting point C to point B.
Using your compass and straight edge transfer the angle ACB to the endpoint of each of the
previous two segments. One can use properties of similar triangles to conclude that this divides
the line L into 3 equal length segments.
C
A
B
L with length 1
If b is the length of each segment and we define 3 b b b b to be the length gotten
by adjoining these three line segments end to end, then 3 b 1 . Thus b is the multiplicative
1
1
inverse of 3, so b . In general,
corresponds to the length of each segment when the unit
3
m
segment is divided into m equal segments. This gives us a geometric model for our abstract
notion of multiplicative inverse of a positive integer m. We will always have this model in the
back of our mind as we proceed to develop our extended number system. We will refer to it as
our “linear model”.
We know that m 0 0 1 for any integer m. So there does not exist a multiplicative inverse for
the integer 0. However, as indicated above, there does exist a multiplicative inverse for any
positive integer.
31
Copyright: Gary Harris, 2009
Exercises for Section 3-1.
Exercise 3.1.1. Solve the following equations for the value of x. Show all steps stating which
properties are used in each step. Use only what we have described in Section 3.1. In particular
we do not yet have rules (formula or definitions) for adding two fractions or for multiplying two
fractions. We only have the concepts of addition of integers, multiplication of integers, additive
identity, multiplicative identity, additive inverse, and multiplicative inverse.
i. 2 x 3 6
ii. (3 x) 1 2
Exercise 3.1.2. Use your compass and straightedge to divide the given line into three equal
length line segments.
A
B
Exercise 3.1.3. Provide two other (different from our linear model) models for the meaning of
1
the multiplicative inverse of a positive integer. That is, that demonstrates the meaning of
for
m
a positive integer m.
32
Copyright: Gary Harris, 2009
Section 3.2
Multiplication of positive integer fractions
We take the same approach as we did when we added the concept of additive inverse to
our positive integers. We will require the associative and commutative properties of
addition and multiplication, and the property that multiplication distributes over addition.
Then we will see what rules must follow from this requirement. We will start with multiplication
of a positive integer times the multiplicative inverse of a positive integer . Suppose that m and n
are positive integers. Our first definition is inspired by our linear model and our definition of
multiplication of positive integers.
1
For positive integers m and n, we define n to be the length gotten by adding line
m
1
segments of length
together n times.
Notice if we assume associativity, we have
m
1
1
n m n m n 1 n .
m
m
1
This means that n is the length when added to itself m times is equal to n. That is, it is the
m
length of each segment when the length n is divided into m equal length segments. We use the
n
notation to represent this length. Thus we have the definition of the positive integer fraction
m
n
and obtain the following
m
1 n
Definition of integer times a multiplicative inverse: n .
m m
Demonstration for n 4 and m 3.
4
1
1/3
1/3
1
1/3
1/3
1/3
1
1/3
1/3
1/3
33
1
1/3
1/3
1/3
1/3
Copyright: Gary Harris, 2009
Fact 3.2.1. m
n
n for positive integers m and n.
m
m
1
m 1 for any
m
m
positive integer m. This is our first occurrence of having different symbols that represent the
1 2 3
m
same number. For example the “fractions” , , , , , all represent the same number 1.
1 2 3
m
This has significant ramifications which we will eventually have to discuss, but for the time
being we will continue as usual.
Form our definition of integer times a fraction it follows that
Suppose that m, n, and p are positive integers. If we enforce associativity and use our
definition of multiplication by a positive integer, we see that
m
1
1 nm
n n m ( n m)
.
p
p
p
p
m nm
Fact 3.2.2. n
for positive integers m, n, and p.
p
p
We now consider multiplication of a positive integer fraction times a positive integer
1 1
fraction. We begin with for positive integers m and n. Again we will enforce
n m
associativity and commutativity. So
1 1 1
1
(n m) n m 11 1
m
n m n
1 1
So is the multiplicative inverse of n m .
n m
1 1
1
Fact 3.2.3.
for positive integers m and n.
n m nm
m p
for positive integers m, n, p, and q.
n q
1 1
m p
1
1
1
m p
m p (m p) (m p)
n q
n
q
n p n p
n q
Finally we consider
Definition of fraction times a fraction.
m p m p
for positive integers m, n, p, and q.
n q n p
This would have to be the definition for a fraction times a fraction if we assume
commutativity and associativity. But let’s now just assume this is our definition and see if we
can derive commutativity and associativity directly.
Using the definition of multiplication of fractions and commutativity of multiplication of
integers we obtain
34
Copyright: Gary Harris, 2009
m a m a a m a m
.
n b nb b n b n
So multiplication of fractions is commutative because multiplication of integers is commutative.
Next consider
m a c m a c m (a c )
.
n b d n b d n (b d )
Using the associativity of multiplication of integers we get
m (a c) (m a ) c
.
n (b d ) (n b) d
Again we use the definition of multiplication to get
(m a) c m a c m a c
.
( n b) d n b d n b d
So we see that associativity of multiplication of fractions follows directly from associativity of
multiplication of integers and our definition of multiplication.
Fact 3.2.4. Multiplication of positive integer fractions is associative.
Finally note that
Fact 3.2.5.
n m nm
m
n
1 . So is the multiplicative inverse of .
m n m n
m
n
n
1
.
m m
n
This is why
m
n m 1 mb.
a n a n a
b
b
Exercises for Section 3.2
Exercise 3.2.1. Provide a model that provides a concrete meaning for each of the expressions
i.
2
3
ii.
3
2
1 1
iii.
2 3
2 2
iv.
3 5
35
Copyright: Gary Harris, 2009
v.
1 4
2 3
36
Copyright: Gary Harris, 2009
Section 3-3
Addition of fractions
Now let’s turn our attention to addition. Suppose that m, n, and p are positive integers
m p
and we have defined some meaning for the expression . Consider the following
n n
m p
expression: n . If we require multiplication to distribute over addition, we must have
n n
m
p
m p
n n n .
n
n
n n
Applying Fact 3.2.1 from Section 3.2 we get
m
p
m p
n n n m p.
n
n
n n
m p
In our linear model this would mean that is the length when added to itself n times gives
n n
m p
the length m + p. This is exactly what we mean by
in that model.
n
An alternate way to look at this would be to observe that
m p
1
1
1 m p
.
m p (m p)
n n
n
n
n
n
Here we used the definition of an integer times a fraction and our assumption that multiplication
distributes over addition. Either way we look at it, we see that we must have the
m p m p
Definition of addition of fractions with common denominators:
for positive
n n
n
integers m, n, and p.
37
Copyright: Gary Harris, 2009
Going back to our linear model we can visualize this quite easily. For example consider the case
n4
m
m+ p
p
Now suppose we have positive integers m, n, p, and q. We want to define the addition
m p
q
n
. We know that 1 and 1 . So we can conclude that
q
n q
n
m p q m n p qm n p qm n p
.
n q q n n q qn nq
nq
m p qm n p
n q
nq
Thinking of our linear model we observe that if we divide a line segment of length m (a
positive integer) into 1 line segment then that one line segment has length m. This is what is
m
meant by m
in our linear model.
1
m
Definition of integer as a fraction. m
for any positive integer m.
1
Definition of addition of fractions.
So any positive integer can be represented as a fraction. In particular 1
1
m 1 m 1 m m
for any positive integers m and n.
n 1 n 1 n n
1
and
1
Fact 3.3.1. The integer 1 is the multiplicative identity (on positive integer fractions).
38
Copyright: Gary Harris, 2009
We extend the above definitions for positive integers to include 0.
0
Definition: 0 .
1
This definition is consistent with our linear model in that any line segment of length 0
0
when divided into 1 segment would produce a segment of length 0: 0 . In fact any segment
1
of length 0 when divided into n equal segments would produce n segments each of length 0:
0
0 . Notice that if n is any positive integer then, assuming our definition of multiplication for
n
fractions with positive integer numerators and denominators extends to include fractions with 0
numerator, we have
0 n 0 n0 0
0
,
1 n 1
n
n
which again is consistent with our linear model.
Now assume that we extend our definition of addition of fractions with positive integer
numerators and denominators to include fractions with 0 numerators. We then have
m 0 m 0 n 1 m 0 m m
0
.
n 1 n
1 n
n
n
Fact 3.3.2. The integer 0 is the additive identity (on positive integer fractions).
Note that if m is a positive integer then we have
0 m 0 m0 m
m
m.
1 1 1
1
1
0
is an additive identity for the positive integers. However, we saw in Section 1.2, Fact
1
1.2.4, that there can be only one additive identity for the positive integers. So, in fact, it follows
0
0
that must equal 0. Hence our definition 0 is forced upon us.
1
1
Also note that
m 0 m 0 m 0
0
0.
n 1 n 1 n n
m
Fact 3.3.3. 0 0 for any positive integers m and n.
n
a
m a
m
Next let’s suppose we have two fractions and , and we know that 0 . We
b
n b
n
compute
0 m a m a
1
0
m a
.
1 n b nb
nb
So
So
39
Copyright: Gary Harris, 2009
0 m a
1
.
nb
By equal actions and associativity, we get
1
0 ( n b) m a
( n b) ( m a ) 1 m a .
nb
Since we know that 0 (n b) = 0, it follows that m a 0
From Section 2.1, Fact 2.1.3, we conclude that either m 0 or a 0 (or perhaps both). We have
m a
m
a
Fact 3.3.4. If 0 , then either
0 or 0 .
n b
b
n
A similar argument (See Exercise 3.3.4) can be used to conclude
m
0 m 0.
n
Algebraic Justification of the Key Properties.
Fact 3.3.5.
Our definition of addition of fractions came about under the assumption of
commutativity, associativity, and distributivity of multiplication over addition. Let’s begin with
the definition of addition of fractions and see if we can obtain these properties directly from the
corresponding properties of addition and multiplication of integers.
We begin with commutativity. Consider
m a mb na an bm a m
.
n b
nb
bn
b n
Notice we used commutativity of both addition and multiplication of integers.
Next we consider the distributive property. Consider
m a c m a d b c m (a d b c )
.
n b d n
bd
n (b d )
Here we used the definitions of addition and multiplication. Next we use the distributive property
for the integers in the numerator to get
m (a d b c) m (a d ) m (b c)
.
n (b d )
n (b d )
Now we use commutativity, associativity, and the definition of addition of fractions with
common denominators to get
m (a d ) m (b c) m (a d ) m (b c)
..
n (b d )
n (b d ) n (b d )
Using the definitions of multiplication of fractions and multiplicative inverse and identity, we
conclude
m (a d ) m (b c) m a m c
.
n (b d ) n (b d ) n b n d
Putting this all together we obtain the distributive property for fractions:
40
Copyright: Gary Harris, 2009
m a c m a m c
.
n b d n b n d
Finally we consider associativity of addition of fractions. To keep the algebra from
looking so messy, we will henceforth adopt the common convention that juxtaposition means
multiplication. In other words we will use mn to denote the product m n . We want to show
that
m a c m a c
.
n b d n b d
Let’s start with the term on the left and see what happens as we do the algebra. Using the
definition of addition twice, we obtain
m a c m ad bc m(bd ) n(ad bc)
.
n b d n bd
n(bd )
Using the distributive property for integers
we get
m(bd ) n(ad bc) m(bd ) (n(ad ) n(bc))
.
n(bd )
n(bd )
We now use associativity of addition and multiplication of integers to get
m(bd ) (n(ad ) n(bc)) (m(bd ) n(ad )) n(bc) ((mb)d (na)d ) (nb)c
.
n(bd )
n(bd )
(nb)d
Using the distributive property for integers (factor out the d), and the definition of addition of
fractions we get
((mb)d (na)d ) (nb)c mb na c m a c
.
(nb)d
nb d n b d
A subtle issue is raised with our definitions of addition and multiplication of fractions.
a
We have observed above that for any positive integer a we have 1 , so it follows that for any
a
positive integers m and n we have
m a m am
.
n a n an
Thus all fractions in each the following two sets are equal to each other,
n2
n 1
1 2 3
2 4 6
, and , , , ,
, .
, , , ,
n3
n5
3 6 9
5 10 15
The question begged is “Do our definitions of multiplication and addition actually depend on our
choice of which fractions from each set we choose to use?” For example, does
2 2 10 2 3 2 26
3 10
3 10
30
produce the “same fraction” as does
41
Copyright: Gary Harris, 2009
6 3 15 6 9 3 117
?
9 15
9 15
135
26 2 13 2 13 13
117
26
the “same fraction” as
? Well, we have
and
30 2 15 2 15 15
135
30
117 9 13 9 13 13
117
26
and
are the “same fraction” in the sense that they are
. So
135 9 15 9 15 15
135
30
13
both equal
. We see that we will be forced to consider exactly what it means for two very
15
different looking fractions to actually be the “same fraction”. For the time being we will
postpone this discussion and continue with our focus on the algebraic structure.
So is
Exercises for Section 3.3.
1 3 4
Exercise 3.3.1. Provide two models for the expression 2.
2 2 2
m2 mn m
1.
mn
n
(You must show all your steps, indicate which property you used at each step.)
Exercise3.3.2. Suppose that m and n are positive integers. Show that
Exercise 3.3.3. Provide model showing why we would not want to define addition of fractions
m p m p
by
.
n q nq
Exercise 3.3.4. Suppose that m and n are integers and n 0. Show that
42
m
0 m 0.
n
Copyright: Gary Harris, 2009
Section 3.4
Fractions with negative integers.
We are now ready to extend our structure to include negative integers. Recall that the
symbol –m denotes the additive inverse of the integer m and is defined by the equation
m (m) 0.
Definition of additive inverse of a fraction. For positive integers m and n,
additive inverse of
m
will denote the
n
m
m
m m
and is defined by the equation
is the thing we
0 . So
n
n
n n
m
to get 0.
n
We extend our definitions of multiplication and addition to include negative integers and
observe the following result:
m m m (m) 0
0.
n n
n
n
m
m
This tells us that
is the additive inverse of .
n
n
add to
Similarly we compute
m m ( n ) m n m ( n) n m
0 m
0
0.
n n
n ( n)
n (n)
n (n) n (n)
m
m
This tells us that
is also the additive inverse of . Thus we have
n
n
m m m
Fact 3.4.1.
.
n
n
n
m
m
m
. But we know that the
n
n
n
m
m
additive inverse of the additive inverse of
is
itself. So we have
n
n
m m
Fact 3.4.2.
.
n n
Consider the following calculation:
Since 0
m
0 for any integers m and n, it follows that there are no integers so that
n
m
1 . Hence 0 is the only number for which there is no multiplicative inverse.
n
m
The set all fractions
for integers m and n with n 0 , with addition, +, and multiplication, ,
n
is a commutative ring with unit, with the added property that every element except for the
additive identity has a multiplicative inverse. It is called the rational number field.
0
43
Copyright: Gary Harris, 2009
In summary.
We can use the algebraic structure of the field of rational numbers to solve for the value of x
equations like following: mx n q . Let’s look at a particular example.
Solve for the value of x if 2 3x 7 .
-2+(2+3x) = -2+7 . equal actions
(-2+2) +3x = 5 .
associativity of addition
0+3x = 5 .
definition of additive inverse
3x = 5 .
definition of additive identity
1
1
equal actions
(3x) 5 .
3
3
1
1
Associativity of multiplication
3 x 5 .
3
3
5
1
definition of multiplication
3 x
3
3
5
definition of multiplicative inverse
1 x .
3
5
definition of multiplicative identity
x .
3
Exercises for Section 3.4.
Exercise 3.4.1. Solve the equation 3 4 x 2 for the value of x. Show each step and indicate the
property you used in each step.
Exercise 3.4.2. Verify that m
Exercise 3.4.3. Verify that
n
m n
.
p
p
n p
n p
.
m q
mq
(m3n mn)n
.
Exercise 3.4.4. Simplify the expression
mn
Exercise 3.4.5. Solve the following equation
for the value of x. Show each step and indicate the property you used in each step.
44
Copyright: Gary Harris, 2009
Chapter 4: Representations of Fractions
Introduction.
In Chapter 3 we completed our development of the Rational Number Field with all the
requisite algebraic structure. We now turn our attention to the representation of rational
numbers, in particular the conversion of a fractions to decimal representations, and the
conversion of decimal representations into corresponding fraction representations. The
conversion of fractions to decimal representations will involve repeated application of the
Division Algorithm, and the conversion from decimals to corresponding fraction representations
will involve use of the Geometric Series and the concept of limiting value. The Chapter 4
concept map outlines this progression.
Chapter 4. Representations of Fractions
The Rational Number Field
DA: Division Algorithm
Rational numbers
Fractions
Rational Numbers
Decimals
Finite decimals
Infinite repeating
decimals
Equivalence
GS: Geometric Series
45
Copyright: Gary Harris, 2009
Section 4.1
Equivalence of fractions
We begin with a very formal calculation. Let
m
a
and
be two integer fractions.
b
n
Suppose that
m a
.
n b
Then by equal actions we have
m
a
(nb) (nb) .
n
b
By associativity, commutativity, and the definition of multiplication we get
1
1
n (mb) b (na) .
n
b
By multiplicative inverses and multiplicative identity we have
mb na .
On the other hand suppose we know that
mb na .
Then by equal actions we have
1
1
(mb)
(na).
nb
nb
By associativity, commutativity, and definition of multiplication we get
m1 a1
b n .
n b bn
By multiplicative inverses and multiplicative identity we have
m a
.
n b
So we have formally obtained the following
Formal Definition of Equivalence of Fractions.
m a
if and only if mb na .
n b
At this point we could simply accept this as our formal definition of the equivalence for
m
a
the two fractions
and . However we have already encountered a notion of equivalent
b
n
fractions. Recall we observed in Section 3.3 that all fractions in each the following two sets are
equivalent to each other,
2n
n
2 4 6
1 2 3
, , , , , and , , , , , .
3n
5n
3 6 9
5 10 15
3
20
4 6
For example and
. In both cases we see that the “cross multiplication” process
15 100
6 9
in the formal definition does work: 4 9 6 6 and 3 100 15 20 . Notice that all the fractions
46
Copyright: Gary Harris, 2009
2
2n
n
1 2 3
2 4 6
in , , , , , are equivalent to and all the fractions in , , , , , are
3
3n
5n
3 6 9
5 10 15
1
2n
2 4 6
equivalent to . We can observe that any two fractions in , , , , , are equivalent
5
3n
3 6 9
2
n
1 2 3
by virtue of them both being equivalent to
and any two fractions in , , , , , are
3
5n
5 10 15
1
2
equivalent by virtue of them both being equivalent to . We will think of as being the
5
3
1
2n
2 4 6
“reduced form” for all the fractions in , , , , , and as being the “reduced form”
5
3n
3 6 9
n
1 2 3
for all the fractions in , , , , , . So we have a second notion of equivalence of
5n
5 10 15
fractions; namely,
Observed Definition of Equivalence of Fractions:
m a
m
a
and
have the
if and only if
n b
b
n
same reduced form.
a
m
and
having the same reduced form
b
n
m
a
imply mb na , and why does mb na imply
and
have the same reduced form? In other
b
n
words, are our formal definition and our observed definition the same concepts?
This begs the following question: Why does
Suppose m and n are non-negative integers and n 0 . Let gcd(m, n) denote the largest
integer that is a factor of both m and n. So gcd(m, n) is a positive integer that divides evenly
(with remainder 0) into both m and n and is the largest integer that does so. Thus we have
ˆ for some positive integer m̂ , and n gcd(m, n) nˆ for some positive integer n̂
m gcd(m, n) m
. Hence
m gcd(m, n) mˆ gcd(m, n) mˆ mˆ
n gcd(m, n) nˆ gcd(m, n) nˆ nˆ
ˆ , nˆ) 1.
and m̂ and n̂ have no common integer factors other than 1, that is to say that gcd(m
Definition of Reduced Form of a Fraction: We call the fraction
fraction
mˆ
the reduced form of the
nˆ
m
.
n
2
4 6
because they both have the same reduced form, namely . However
3
6 9
4 20
4 2
20 5
.
because the reduced form for is , but the reduced form for
6 36
6 3
36 9
For example
47
Copyright: Gary Harris, 2009
Suppose
We let
Thus
m a
by our observed definition. That is, they have the same reduced form.
n b
mˆ
be the reduced form of both. So
nˆ
m gcd(m, n) mˆ
a gcd(a, b) mˆ
and
.
n gcd(m, n) nˆ
b gcd(a, b) nˆ
ˆ )(gcd(a, b)nˆ) gcd(m, n) gcd(a, b)mn
ˆˆ
mb (gcd(m, n)m
and
ˆ ) gcd(m, n) gcd(a, b)nm
ˆˆ.
na (gcd(m, n)nˆ)(gcd(a, b)m
We see that mb na as integers. So
m a
by the formal definition.
n b
m
a
and
are equivalent by the formal definition. So we know that
b
n
m
a
mb na as integers. Does this imply that
and
must be equivalent by our observed
b
n
m
a
definition? That is, must
and have the same reduced form?
b
n
ˆ
a
mˆ
m
a
Let
and
be the reduced fraction forms of
and
, respectively. Then
ˆ
b
nˆ
n
b
a gcd(a, b) aˆ
m gcd(m, n) mˆ
and
.
b gcd(a, b) bˆ
n gcd(m, n) nˆ
Now suppose the
Also m̂ and n̂ have no common factors other than 1, and â and b̂ have no common factors other
than 1. Since mb na we have
ˆ )(gcd(a, b)bˆ) (gcd(m, n)nˆ)(gcd(a, b)aˆ ) .
(gcd(m, n)m
Applying the associative and commutative properties of multiplication we have
ˆ ˆ) (gcd(m, n) gcd(a, b))(na
ˆ ˆ) .
(gcd(m, n) gcd(a, b))(mb
Multiplying both sides of this equation by the multiplicative inverse of gcd(m, n) gcd(a, b) , we
get
ˆ ˆ na
ˆˆ.
mb
Recall that m̂ and n̂ have no common factors other than 1, and â and b̂ have no common factors
ˆ ˆ are equal, all the integer factors of one must also
other than 1. Since the two integers m̂bˆ and na
be integer factors of the other. None of the factors of m̂bˆ due to m̂ can come from n̂ , so all the
ˆˆ.
integer factors in m̂bˆ due to the presence of m̂ must be coming from â in the product na
48
Copyright: Gary Harris, 2009
ˆˆ
Likewise all the factors of m̂bˆ due to the presence of b̂ must be coming from n̂ in the product na
m
a
ˆ aˆ and nˆ bˆ. Hence
. So we must have m
and have the same reduced form, so they are
b
n
equal by our observed definition.
Fact 4.1.1. Our formal definition and our observed definition of equivalence of fractions are
m a
m
a
exactly the same. That is,
and have
if and only if mb na as integers, if and only if
n b
b
n
the same reduced form.
This gives us a very easy way to tell if two fractions are equal as fractions.
20
55
20 55
since 20 99 1980 and 36 55 1980 . Also notice that
and
have
36
36 99
99
20 190
5
the same reduced form; namely, . However
since 20 360 7200 , but
36 360
9
190
19
20 5
36 190 6840 . Notice that the reduced form for
is , but the reduced form for
is
360
36 9
36
.
Examples:
Recall the definition of fraction multiplication:
m a ma
m M
. Suppose
and
n b nb
n N
M A MA
a A
as fractions. Would the product
give a fraction equal to the first product?
N B NB
b B
Let’s write down the equalities we know from our assumption.
mN nM and aB bA .
Does this force
ma MA
?
nb NB
Well, let’s try the cross multiplication and see if we get the same thing:
(ma)( NB) (mN )(aB) (nM )(bA) (nb)(MA) .
The first equality is because of associativity and commutativity, the second is by substitution and
the last again by associativity and commutativity.
This checks, so the two products yield equal fractions.
Fact 4.1.2. Multiplication is well-defined. That is, it is independent of the choice of the forms to
represent the fractions being multiplied. The products will again be equal fractions.
49
Copyright: Gary Harris, 2009
m a mb na
m M
a A
. Suppose
and as
n b
b B
n N
nb
MB NA mb na
fractions. Does it follow that
as fractions? We compute the cross product.
NB
nb
Does
(MB NA)(nb) ( NB)(mb na) ?
Recall the definition of addition:
We have the associative, commutative, and distributive properties and the identities
mN nM and aB bA .
The algebra is a bit tedious, but gives us a good opportunity to hone our skills and use the
structure we have been studying.
(MB NA)(nb) (MB)(nb) ( NA)(nb) ………..Distributive property
(nM )(bB) (nN )(bA) …. . . . Assoc. & comm.
(mN )(bB) (nN )(aB) ….. . . . Substitution
(mb)( NB) (na)( NB) ............ Assoc. & comm.
( NB)(mb na) …………….. Distributive property
Well, that wasn’t so bad.
Fact 4.1.3. Addition is well-defined. That is, it is independent of the choice of the forms to
represent the fractions being added. The sums will again be equal fractions.
An example of another possible definition for addition is given by the following:
Suppose we were to try to define addition as follows:
m a ma
.
n b nb
2 4 6
2 4
4
With this “definition” we would get the answer . We know that . Using
in
3 5 8
3 6
6
4 4 8
2
6
8
the place of we get . But we easily see that and
are not equal as fractions.
6 5 11
11
8
3
Thus this definition is not well-defined. It depends on the choice of the representation we use for
the fractions involved.
Another problem with this “definition” is that it is contrary to our original meaning of
1
addition with common denominator, and in fact, with our basic definition of the meaning of .
n
1 1
For example, we know from our basic linear model that 1 , however the above
2 2
1 1 2
2 1
“definition” would yield
, which is very troublesome since . Thus it follows
2 2 4
4 2
1 1 1
1
that 0 1 0, which we know is not true.
2 2 2
2
50
Copyright: Gary Harris, 2009
Exercises for Section 4.1.
Exercise 4.1.1. For each of the following pairs of fractions, determine which are equivalent.
First use the formal definition. Then use the observed definition by finding the reduced forms of
each fraction and comparing them.
1.
18
27
and
54
81
2.
56
72
and
189
243
3.
320
384
and
448
512
Exercise 4.1.2. Show that
14 12 12 15
.
21 20 18 25
Exercise 4.1.3. Show that
14 12 12 15
.
21 20 18 25
51
Copyright: Gary Harris, 2009
Section 4-2
Decimal representation of integer fractions (rational numbers)
Finite and repeating decimals
Recall that the decimal expansion of a number is based on powers of
0.45
4
5
1 1
4 5
10 100
10 10
1
. Examples:
10
2
2
3
4
5
0
0
3
2
5
1 1
1
1
1
0.00325
0 0 3 2 5
10 100 1000 10000 100000
10 10
10
10
10
Notice in each case the given “finite decimal” can easily be written as the corresponding integer
fraction:
2
40
5
45
1 1
0.45 4 5
10 10 100 100 100
3
4
5
5
5
5
325
1
1
1
1
1
1
0.00325 3 2 5 300 20 5
10
10
10
10
10
10 100000
So a general finite decimal can always be converted to the corresponding integer fraction:
d
d
d1 d2
2 33 nn
10 10 10
10
n 1
n2
n 3
d (10) d 2 (10) d3 (10) d n d1d 2 d3 d n
1
.
10n
10n
0.d1d 2 d3 d n
So every finite decimal yields an integer fraction. Does every integer fraction yield a finite
decimal?
Division Algorithm (DA)
Suppose m and n are positive integers with n m . Recall that m represents the number
of objects in all the collections containing m objects and n represents the number of objects in all
collections containing n objects. Since n m, we can take n objects away from any of the sets
containing m objects, leaving a set containing m n objects remaining. If m n n then we can
not take out n objects from the remaining set containing m n objects. However, if n m n we
can again take out n objects, leaving a set containing (m n) n m 2n objects remaining. The
idea is to continue taking out n objects at a time until the remaining set has fewer than n objects.
Division Algorithm (DA): If m and n are positive integers with n m , then there exist nonnegative integers q and r so that
m qn r and 0 r n
52
Copyright: Gary Harris, 2009
The integer q is called the quotient, and the integer r is called the remainder. We say n goes into
m “q times with remainder r”.
Examples:
8 2 4 0
(4 goes into 8 two times with remainder 0)
14 1 4 10 2 4 6 3 4 2
(4 goes into 14 three times with remainder 2)
Question: What does this have to do with decimal expansions for integer fractions?
Consider the fraction
14 3 4 2 3 4 2
2
3 .
4
4
4 4
4
14
. From the DA we have
4
Since the remainder 2 is less
than the number 4, we can not
take 4 objects from any collection
of 2 objects. However we can
take 4 objects from a collection
of 20 objects.
So let’s consider the following:
2 2 10 20 1
.
4 4 10 4 10
We apply the division algorithm
with m = 20 and n = 4.
20 5 4 0
So
20 5 4 0 5 4 0
4 4
5 0 5.
4
4
4
4
4 4
So
2 20 1
1
5 0.5.
4 4 10
10
Hence
14
2
3 3 0.5 3.5.
4
4
Let us practice using the division algorithm to convert the following integer fractions to decimal
form:
3
Example1:
8
53
Copyright: Gary Harris, 2009
Step 1
a) We can not take any 8’s out of 3 but we can take 8’s out of 30.
3 30 1
8 8 10
b) Apply the DA with m = 30 and n = 8.
30 3 8 6
(Eight goes into 30 three times with remainder 6.)
c) Use substitution, the definition of addition, and the distributive property of multiplication
3
over addition to compute the value of .
8
3 30 1 3 8 6 1 3 8 6 1
1 6 1
3
8 8 10 8 10 8 8 10
10 8 10
Step 2
a) repeated with the remainder from step1 part b) being divided by the same number 8.
We can not take any 8’s out of 6 but we can take 8’s out of 60.
6 60 1
8 8 10
b) repeated. Apply the D.A. with m = 60 and n = 8.
60 7 8 4
(Eight goes into 60 seven times with remainder 4.)
c) repeated. Use substitution, the definition of addition, and the distributive property of
6
multiplication over addition to compute the value of .
8
6 60 1 7 8 4 1 7 8 4 1
1 4 1
7
8 8 10 8 10 8 8 10
10 8 10
6
d) Substitute this value for into the result of step 1 part c) and use the distributive property of
8
3
multiplication over addition to compute the value of .
8
3
1 6 1
1 1 4 1 1
1 1 4 1
3 3 7 3 7
8
10 8 10
10 10 8 10 10
10 10 8 10
Step 3
a) repeated with the remainder from step2 part b) being divided by the same number 8.
We cannot take any 8’s out of 4 but we can take 8’s out of 40.
4 40 1
8 8 10
2
b) repeated. Apply the D.A. with m = 40 and n = 8.
40 5 8 0
(Eight goes into 40 five times with remainder 0.)
54
2
Copyright: Gary Harris, 2009
c) repeated. Use substitution, the definition of addition, and the distributive property of
4
multiplication over addition to compute the value of
.
8
4 40 1 5 8 0 1 5 8 0 1
1
5
8 8 10 8 10 8 8 10
10
4
d) substitute this value for
into the results from step 2 part d) and use the distributive property
8
3
of multiplication over addition to compute the value of .
8
2
2
2
2
2
3
3
1 1 4 1
1 1 1 1
1 1
1
3 7 3 7 5 3 7 5 0.375
8
10 10 8 10
10 10 10 10
10 10
10
m
into decimal form notice that
n
each step of the process involves dividing the remainder in part b) from the previous step by the
same integer n. So if we get remainder of 0 at any step in the process, then the process stops and
m
we have converted
into a “finite decimal” expression.
n
Observation: When applying DA to convert the integer fraction
Example 2:
2
3
Step 1
a) We cannot take any 3’s out of 2 but we can take 3’s out of 20.
2 20 1
3 3 10
b) Apply the DA with m=20 and n=3.
20 6 3 2
(3 goes into 20 three times with remainder 2)
c) Use substitution, the definition of addition, and the distributive property of multiplication
2
over addition to compute the value of .
3
2 20 1 6 3 2 1 6 3 2 1
1 2 1
6
3 3 10 3 10 3 3 10
10 3 10
Step 2
a) repeated with the remainder from step1 part b) being divided by the same number 3.
We cannot take any 3’s out of 2 but we can take 3’s out of 20.
2 20 1
3 3 10
b) repeated. Apply the DA with m=20 and n=3.
20 6 3 2
55
Copyright: Gary Harris, 2009
(3 goes into 20 three times with remainder 2)
c) repeated. Use substitution, the definition of addition, and the distributive property of
2
multiplication over addition to compute the value of .
3
2 20 1 6 3 2 1 6 3 2 1
1 2 1
6
3 3 10 3 10 3 3 10
10 3 10
2
d) Substitute this value for
into the result of step 1 part c) and use the distributive property of
3
2
multiplication over addition to compute the value of .
3
2
1 2 1
1 1 2 1 1
1 1 2 1
6 6 6 6 6 At
3
10 3 10
10 10 3 10 10
10 10 3 10
this point we observe that each step is just going to continue giving the same remainder of 2, and
2
1
yielding the same expression for , with another power of showing up at each consecutive
3
10
step. It would appear that
2
3
2
1 1
1
6 6 6 0.666
3
10 10
10
2
That is to say, the decimal expansion of has 6’s repeating forever, whatever “forever” means.
3
2
2
Question: What does a decimal expression that repeats “forever” actually mean?
Of course we know what any finite string means, just add them up. For example, if n is a
positive integer then
666 6
1 1
1
1
0.666 6 6 6 6 6
,
10n
10 10
10
10
2
3
n
where there are n sixes in the decimal on the left and n sixes in the numerator on the right. But
what does it mean to “go on forever”?
Try one more example.
11
Example 3:
(Exercise in class)
90
m
for positive integers m
n
and n, each step involves the remainder from the previous step being divided by the same
integer n. Since the remainder, when dividing by n, must be an integer r with 0 r n there are
only n possible values for the remainder r. Thus, if we keep doing the step-by-step process
indicated in the examples above, we must eventually get either 0, or repeat a non-zero remainder
we have gotten before. (A remainder must be 0 or repeat within at most n steps.) Remember that
the process involves dividing each previous remainder by the same integer n. Hence, we either
Observation: When using the DA to find the decimal expansion of
56
Copyright: Gary Harris, 2009
have a finite decimal, or all the digits we got for the previous repeated remainder will be gotten
1
again, with the powers of growing proportionally. Thus we get a decimal expansion that
10
has some pattern of digits repeating “forever”.
Major Observation: Every integer fraction must correspond to a finite decimal or an infinitely
repeating decimal expansion.
Again we have the question: What does a decimal expression that repeats “forever” actually
mean? The answer to this question is given in Section 4.3.
Exercises for Section 4.2
Exercise 4.2.1. Find the decimal expansions for the following fractions:
a.
b.
51
200
53
225
Exercise 4.2.2. Explain why an integer fraction must have either a finite decimal representation
of an infinitely repeating decimal expression.
Exercise 4.2.3. Explain why an integer fraction has a finite decimal expansion if and only if the
denominator of the reduced form of the fraction has only factors of 2 and/or 5.
57
Copyright: Gary Harris, 2009
Section 4-3
Converting decimal to fractions
Geometric Series
As we saw in Section 4-2, any finite decimal can easily be converted into the equivalent
integer fraction. The question we want to address here is “What about decimals that are not
finite?” Let’s look a familiar example of a decimal expansion that repeats “forever”.
Example: 0.999999 which is often denoted as 0.99 . Our question is “What does this really
mean?”
Let’s look at what we do know:
1
0.9 9
10
1 1
0.99 9 9
10 10
2
2
1 1
1
0.999 9 9 9
10 10
10
2
3
3
1 1
1
1
0.9999 9 9 9 9
10 10
10
10
4
In fact if N is a positive integer then
2
3
4
N
1 1
1
1
1
0.9999 9 9 9 9 9 9 .
10 10
10
10
10
N 9s3
Using the distributive property of multiplication over addition, we see that
1
2
3
N 1
9 1 1 1
1
0.9999 9
1
10 10 10 10
10
N 9s
1
Notice that each term in the sum is times the previous term. There is a name for such
10
sums. Let r be any positive number. The summation of the form
1 r r2 r3 r4
is called a geometric series and the number r is called the ratio of the series. In our case
We let S N denote the sum of the first N+1 terms of this series:
SN 1 r r 2 r 3 r 4 r N .
58
Copyright: Gary Harris, 2009
9
9
1 S0
10
10
9 1 9
0.99 1 S1
10 10 10
So 0.9
2
9 1 1 9
0.999 1 S2
10 10 10 10
2
3
9 1 1 1 9
0.9999 1 S3
10 10 10 10 10
1
2
3
N
9 1 1 1
1 9
0.9999 9
1 S N
10 10 10 10
10 10
( N+1) 9s
So we are looking for a formula for S N .
There is a really cool way to get a formula for S N for any positive number r.
Observe that
rS N r (1 r r 2 r 3 r 4 r N ) r r 2 r 3 r 4 r N 1.
So
S N rS N (1 r r 2 r 3 r 4 r N ) (r r 2 r 3 r 4 r N 1 )
S N rS N 1 (r r ) (r 2 r 2 ) (r 3 r 3 ) (r N r N ) r N 1
S N rS N 1 r N 1
(1 r )S N 1 r N 1
Under the assumption that r 1 we can multiply both sides of this expression by the
multiplicative inverse of ( 1 r ) and conclude
1 r N 1
SN
(1 r )
1
Applying this formula for r we get
10
N 1
N 1
1
1
1
1
N 1
10 1
10
10
Formula 4.3.1 .
SN
1 .
1
9 10
9
1
10
10
59
Copyright: Gary Harris, 2009
Thus
0.9999 9
N 1
N 1
9
9 10 1
1
S N 1 1
10
10 9 10
10
(N+1) 9s
What happens to this expression if we keep putting more and more 9s in the decimal expansion?
In other words, what happens as the value of N gets bigger and bigger?
1
The number
10
N 1
1
keeps getting closer and closer to 0, in fact we can make
10
N 1
as close to
N 1
1
0 as we want by just taking larger and larger values for N. So the number 1 gets closer
10
and closer to the number 1, in fact as close as anyone might want, as N gets bigger and bigger.
N 1
1
We say that the limiting value of 1 is equal to 1 as the value of N is allowed to grow
10
without limitation. That is to say, if we let N increase “forever” we get “infinitely close” to 1.
This is what we mean by
0.999999 1.
9s “forever”
Use the same process to determine the integer fractions represented by the following
“infinite” decimal expansion:
Example 1: 0.55
2
3
1 1
1
0.55 5 5 5
10 10
10
For any positive integer N we have
2
3
N
1
2
N 1
5 1 1
1 1
1
1
1 5
0.555 5 5 5 5 5 1 S N 1.
10 10 10
10 10
10
10
10 10
N 5s
Substituting the value of S N 1
10 1
1
9 10
0.555 5
N
from 4.3.1 we get
N
N
5
5 10 1 5 1
S N 1 1 1 .
10
10 9 10 9 10
N 5s
If we let the decimal go on forever, that is let N grow without limitation, we get
5
5
0.55 1 .
9
9
Major Observation: Any infinitely repeating decimal can be written as an integer fraction.
60
Copyright: Gary Harris, 2009
Conclusion: A number is rational if and only if it can be represented by either a finite decimal or
an infinite decimal with a repeating pattern in its digits.
Question: What can we say, if anything, about an infinite decimal that has no repeating pattern
in its digits, something like 0.d1d 2 d 3 d 4 for infinitely many random digits d j ? Does such a
decimal represent anything? This is the main topic of discussion in Section 5.
Final note on conversion from decimal to fraction form.
Here is an oft used process for converting infinite repeating decimals to fraction form. It
shows up in text books used of pre-service teacher math courses as well as middle school math
textbooks. I contend that it is a good example of a process being substituted in the place of
conceptual understanding that ultimately begs some serious questions. We will apply this process
to a particular example.
Consider the expression 1.111111….. (the ones repeating forever.)
A 1.1111.....
0)
10 A 11.1111......
1)
10 A A 10
2)
3)
(10 1) A 10
9 A 10
4)
10
5)
A
9
10
Our conclusion is that 1.1111 .
9
Now let’s analyze this process step by step, just to be sure we know exactly what is going on.
0) We are considering 1.1111….. to be an object (number) and are assigning the letter A to
represent it. Writing out the expansion using our meaning for decimal representation we
1
1
1
1
get A 1.1111.... 1 2 3 4 .
10 10 10 10
1) We are multiplying our object, A, by 10 (on the left) and using the distributive property of
multiplication to get its equivalent representation on the right:
1
1
1
1
1
1
1
10 A 10 1 2 3 4 10 1 2 3 11.1111
10 10 10
10 10 10 10
2) We are using term by term subtraction to compute 10 A A 10
3) We are using the distributive property to factor out our object A.
4) We compute 10-1=9.
1
5) We multiply both sides of the equation by .
9
Let’s look at another example. Consider the expression 1 10 102 103 104 . Now
apply the same process as above, using exactly the same steps.
0) A 1 10 102 103 104
1) 10 A 10(1 10 102 103 104 ) 10 102 103 104 105
61
Copyright: Gary Harris, 2009
2) 10 A A (10 102 103 104 105 ) (1 10 102 103 104 ) 1
3) (10 1) A 1
4) 9 A 1
1
5) A .
9
Our conclusion is that the same process we did before now yields
1
1 10 102 103 104 . This is utter nonsense. So what is going on?
9
0) We are assuming that 1 10 102 103 104 is an object (number). But it is not a
number. For any positive integer N we apply our previous formula to get
1 10 N 1
. Notice that as the integer N gets bigger and
1 10 102 103 104 10 N
1 10
bigger, the number 10 N 1 just keeps getting bigger and bigger without any upper limit.
So, as the integer N gets bigger and bigger the number 1 10 102 103 104 10N ,
keeps getting bigger and bigger without any upper limit. Some books would say that
1 10 102 103 104 is infinitely large and use the notation
1 10 102 103 104 . So thinking of 1 10 102 103 104 as an object
(number) which follows the usual rules of algebra is questionable from the get go.
1) We are assuming that multiplication distributes over the infinite sum
1 10 102 103 104 . This would be dubious even if we could think of
1 10 102 103 104 as representing a number.
2) We are assuming that subtraction of two infinite sums can be done by subtracting term by
term. Again this would be dubious even if we could think of each infinite sum as
representing a number.
3) We are assuming A is an object that can be factored out, but in this case it is not.
4) We subtract 1 from 10. OK, this we can do.
5) We are treating A as an object (number) on which we can do algebra, but this is not the
case here.
The question is “Why did this process seem to work in the first example and not in the
second?”
N 1
1
1
1
1
1
1
1
10
In the first example we have 1 2 3 4 N for any integer N.
1
10 10 10 10
10
1
10
1
So as the value of N gets bigger and bigger (grows without bounds) the value for
10
1
1
10
closer and closer to 0 (gets “infinitely close to 0”) and the expression
1
1
10
62
N 1
gets
N 1
gets infinitely
Copyright: Gary Harris, 2009
close to the number
1
10
. So in this case it turns out to be OK to think of 1.1111…. as an
9
1
10
object and represent it by the letter A.
1
1
1
10
A 1.1111 lim 1.1111 lim
N
N
1
1
10
N 1’s
N 1
1
1
1
10
10
9
It is a theorem from advance calculus that multiplication distributes over the infinite
addition in such a series, and that the subtraction of two such series can be done in a term by
term manner.
The final conclusion is that the process used above for converting decimals to fractions
uses very deep mathematics that involves the special properties of the geometric series
1
1
1
1
1 2 3 4 . Showing this process works to convert decimals to fractions
10 10 10 10
would be a good exercise in a graduate level real analysis class. I question its use in a preservice elementary teacher class, let alone in an elementary classroom.
Exercises for Section 4.3.
Exercise 4.3.1. Find the integer fraction representation of 0.122.
Exercise 4.3.2. Find the integer fraction representation of 0.1313.
63
Copyright: Gary Harris, 2009
Chapter 5: Other kinds of numbers?
Introduction.
In Chapter 3 we finished our development of the Rational Number Field which consists
of all possible integer fractions. In Chapter 4 we saw that any rational number (integer fraction)
is either equal to a finite decimal, or is the limiting value of an infinite decimal with a repeating
pattern. So we could equally define the Rational Number Field to consist of all possible finite or
infinitely repeating decimals. This seems to leave the door open for the existence of other kinds
of numbers; namely, numbers defined by the limiting values of infinite decimals that have no
repeating patterns. In this chapter we will see that the existence of such numbers is a
consequence of something called the Least Upper Bound Principle. This process is outlined in
the following concept map.
Chapter 5. Other kinds of numbers
The Real Number Field
Possible “other numbers”
Rational Number Field
Numbers as finite or
repeating decimals
Examples
Examples: √ ;
√ integer
√ ⁄ integer
(
and with
not both perfect
Least
Upper Bound
and prime ;
, positive integers
)
, and
powers.
Principle
Irrational Numbers
Numbers which are limiting values of non-repeating infinite decimals
64
and
Copyright: Gary Harris, 2009
Section 5.1
Square Root of 2
2
m
Theorem 5.1.1. There do not exist integers m and n so that 2 .
n
Proof (by contradiction).
m gcd(m, n)mˆ gcd(m, n) mˆ mˆ
Suppose m and n are integers and n 0. Then
for
n gcd(m, n)nˆ gcd(m, n) nˆ nˆ
2
2
mˆ
m
ˆ , nˆ) 1. Assume 2 . Thus 2. So let’s forget about
integers m̂ and n̂ with gcd(m
nˆ
n
(
)
the hat business and simply assume that
for the original m and n.
(1)
m m m m m2
m
2 2.
n n
nn
n
n
Multiplying the last equality in (1) by
yields
(2)
m2 2n2 .
So m 2 is even. We know that the square of an odd integer must again be an odd integer. Thus it
follows from (2) that m must be even. So m 2k for some integer k. Substituting 2k for m in
(2) yields
2
2k 2n2 .
2
But we know that 2k 2k 2k 2 2 k k 4k 2 .
(3)
2
So from (3) we have
(4)
4k 2 2n2 .
Multiplying both sides of (4) by the multiplicative inverse of 2 and using associativity of
multiplication, we see that (4) implies
2k 2 n ,
(5)
which in turn implies that n 2 is even, so n must also be even. However this would imply that m
and n are both divisible by 2 which contradicts the fact that gcd(m, n) 1. So our assumption
2
2
m
that 2 must be false. Hence there is no rational number (integer fraction) whose square
n
is equal to 2.
Question: Why should we believe there is anything (any number) whose square is 2?
One possible reason comes from the Pythagorean Theorem. We can use a compass and straight
edge to construct a right triangle with two sides of unit length (length 1). See Figure 1. Let c
denote the length of the hypotenuse. The Pythagorean Theorem tells that that
c2 12 12 2
Surely such a length (number) c exists, after all we just constructed it.
65
Copyright: Gary Harris, 2009
Figure 1.
c
1
1
Another possible reason is our belief that we should be able to draw a straight line
without lifting our pencil off the paper. That is, draw a line without any holes in it. The word
“continuous” is used in this context. Consider the two sets of positive numbers
L a 0 such that a 2 2 and R a 0 such that 2 a 2
0
a2 < 2
2 < a2
L
R
Surely every positive number must be contained either in L or in R because for any number a,
either a 2 2 or 2 a 2 . If there were no number a so that a 2 2 , then there would have to be a
hole (or gap) in the number line between the set L and the set R. This would make us very
uncomfortable. Also the equation x2 2 0 would have no solution.
This suggests why we think such a number c with c2 = 2 should exist, but why must such a
number exist? This is the question we ponder in the next section.
Recall the Fundamental Theorem of Arithmetic: Every positive integer can be written in
a unique way as the product of prime numbers. Suppose that m is a positive integer, then there is
one and only one way (up to the order of listing the prime factors) to write m as a product of
primes.
m p1r1 p2r2 p3r3 pnrn
So
m2 p1r1 p2r2 p3r3 pnrn
2
p12r1 p22r2 p32 r3 pn2 rn .
The result of this is that the prime factors of m 2 are exactly the same as the prime factors of m.
So, for example, if 3 is a factor of m 2 , then 3 is also a factor of m. Thus the same argument used
66
Copyright: Gary Harris, 2009
2
m
above to prove that there are no integers m and n so that 2 can be used to show there are
n
2
m
no integers m and n so that 3 . For that matter, it can be used to show that for any prime
n
2
m
number p there are no integers m and n so that p .
n
m
In fact we can do a lot better than this. Let be an integer fraction. Suppose that
n
2
m
a
a
m
, equivalently n b .
b
n
2
2
a
a
m
m
We know that if and only if and
have the same reduced form. But
b
b
n
n
2
2
m gcd(m, n)mˆ mˆ mˆ
nˆ 2
n gcd(m, n)nˆ nˆ
2
2
Since all the prime factors of m̂ 2 and n̂ 2 are the same prime factors of m̂ and n̂ respectively and
2
m
ˆ 2 , nˆ 2 ) 1. So we see that the reduced form of is in fact
ˆ , nˆ) 1 , it follows that gcd(m
gcd(m
n
2
mˆ
. The last observation we make is that two fractions in reduced form are equal if and only if
nˆ 2
they have exactly the same numerators and the same denominators.
Our conclusion is the following:
Theorem 5.1.2. The square root of the integer fraction
a
is a rational number if and only if the
b
a
has numerator and denominator that are both perfect squares.
b
Corollary. The only integers with rational square roots are the integers which are themselves
perfect squares.
reduced form of
Proof. Apply Theorem 5.1.2 to the integer m
m
.
1
A careful look at the above discussion should reveal that there is nothing special about
the power 2 in the squaring process. If we replace the power two by any positive integer we
would have the similar conclusion.
67
Copyright: Gary Harris, 2009
Theorem 5.1.3. Suppose that m is a positive integer. It follows that
a
and only if
reduces to a fraction of the form
b
m
a
is a rational number if
b
aˆ m
with gcd(aˆ m , bˆm ) 1.
m
ˆ
b
Exercises for Section 5.1.
Exercise 5.1.1. Show that the square of an odd integer must again be an odd integer.
Hint: Recall that any odd integer can be written in the form 2n + 1 for some integer n.
2
m
Exercise 5.1.2. Show that there are no integers m and n so that 7 .
n
Provide two different arguments.
2
m
Exercise 5.1.3. Show there are no integers m and n so that 28 .
n
Exercise 5.1.4. Show there is no rational number that is equal to
3
8
.
15
Exercise 5.1.3. Show that there is a rational number that is equal to
68
3
16
.
54
Copyright: Gary Harris, 2009
Section 5.2
Irrational numbers
The Least Upper Bound Principle
Let S be a set of numbers. We say that S has is bounded above if there is a number r so that no
number in S is bigger than r. Such a number r is called an upper bound for the set S.
Examples: The set S 1, 2, 4,7 is bounded above. The number 10 is an upper bound for S, so
are the numbers 15, 27, 7.5, and 7. Note that 7 is the smallest number that is an upper bound for
the set S.
{
} is bounded above. The numbers 5, 4, 3, 2.5 are all
The set
upper bounds for the set S. The smallest (least) upper bound for the set S is the number 2. Notice
in this example that the least upper bound for the set S is not actually in the set S.
We agree to place the following (unproved) assumption (axiom) on the set of numbers:
The Least Upper Bound Principle: If a set S of numbers is bounded above, then there exists a
number which is the least upper bound for the set S.
Notice that if we have two different upper bounds for the same set S, then the bigger of
the two cannot possibly be the least upper bound for S since there is an upper bound that is
smaller than it. It follows that the least upper bound for a set must be unique.
Let’s consider some of the consequences of this assumption (axiom).
Let L a 1 such that a 2 2 . We observe the set L is bounded above. For example, 2 is
an upper bound for L. Also is an upper bound for L, but is not an upper bound for L. (See
Exercise 5.2.1.) By the Least Upper Bound Principle there exists a number which is the least
upper bound for the set L. Let c denote this number. So anything in L is less than or equal to c
and c is the smallest such number. That is to say that any number less than c is not an upper
bound for L. This means that, if r is any number less than c, then there is a number in L that is
bigger than r. We claim that it must be the case that c2 = 2. If c2 < 2 then there is a number a
which is bigger than c but still has a2 < 2. Thus there is a number in L that is bigger than c. But
this cannot be the case since c is an upper bound for L. On the other hand if 2 < c2 then there is a
number r that is less than c but still has 2 < r2. Then r is an upper bound for L that is smaller
than c. But this cannot be the case since c is the least (smallest) upper bound for L. The only
possibility left is for c2 = 2. (Again see Exercise 5.2.1.)
Consequence 5.2.1. There exist a number c with c2 = 2. So there exist numbers that are not
rational (integer fractions).
We know that there is no integer that is an upper bound for the set of integers. After all,
if we have an integer N, then N+1 gives us a bigger integer. However this alone does not
69
Copyright: Gary Harris, 2009
exclude the possible existence of some mysterious number that might be bigger than all the
integers. So let’s suppose there is some number that is bigger than all the integers. In other
words, suppose the integers are bounded above. By the Least Upper Bound Principle there
would be a least upper bound for the integers. Let R be this number. Then R – ½ is not an
upper bound for the integers. So there is some integer N so R – ½ < N . It follows that R < N +
½ < N + 1. However this can’t be since N + 1 is an integer and R is an upper bound for the
integers. Thus there cannot be a least upper bound for the integers, hence no upper bound for the
integers.
Consequence 5.2.2. Given any number of any kind, there is an integer bigger than it.
Consider an infinite decimal
2
3
1
1
1
0.d1d 2 d3 d1 d2 d3 .
10
10
10
For any positive integer N, let
2
3
N
1
1
1
1
S N d1 d 2 d3 d N .
10
10
10
10
It will be important to recall that each d j is a digit, 0 d j 9 . Since each digit is non-negative
we see that
#
S1 S 2 S 3 S N S N 1 .
This follows for each positive integer N because we have
N 1
1
S N 1 S N d N 1 .
10
We also note that each d j 9 . Hence for each positive integer N
2
3
N
1 1
1
1
S N 9 9 9 9 0.999 9 1 .
10 10
10
10
So the set S S1 , S2 , S3 , is bounded above. Hence by the Least Upper Bound Principle, S
has a least upper bound. Let r denote the least upper bound of the set S. If is any positive
number, no matter how small, then r r which implies that r is not an upper bound for S.
So there is an element S N in the set S so that r S N . By # this implies that r SM for all
integers M that are bigger than N. Since r is an upper bound for S, we also must have SM r for
all integers M. In summary, we see that no matter how small we choose the positive number ,
there must be an integer N so that
r SM r for all integers M bigger than N.
Recalling the definition of S M we see this means
2
r SM
3
1
1
1
1
d1 d 2 d 3 d M
10
10
10
10
70
M
r
Copyright: Gary Harris, 2009
for all M that are big enough. In decimal form this means that no matter how small the positive
number
r S M 0.d 1d 2 d 3 d M r
.
for all M that are big enough. This is what we mean when we say the infinite decimal
0.d 1d 2 d 3 r . In other words, is the limiting value of all the finite decimals .
Forever
Consequence 5.2.3. Every infinite decimal 0.d 1d 2 d 3 represents some number. The set of all
such numbers constitutes the real number system.
Suppose r is a positive real number, maybe very, very small. Let S be the set of all
numbers of the form nr for some positive integer n:
{ |
}
Suppose that S has an upper bound. Then by the Least Upper Bound Principle there is a number
c that is the least upper bound for S: c is the smallest number so that
Note that r c since r 2r c . The positive number c r is not an upper bound for S, so there
is an integer M so that c r Mr . This implies c Mr r (M 1)r which contradicts the fact
that c is an upper bound for S.
Consequence 5.2.4. For any positive number r, no matter how small, and any number s, no
matter how large, there is an integer N so that Nr > s.
Suppose that r and s are any two irrational numbers and r s . Then there is a positive
integer n so that n(s r ) 1 . Thus nr ns 1. Let m be the least integer that is bigger than ns.
Thus ns 1 m 1 , but m 1 ns since m is the least integer that is bigger than ns. Then it
m 1
follows that nr m 1 ns . So r
s.
n
Consequence 5.2.5. There is a rational number between any two distinct irrational numbers.
Suppose that a and b are any two rational numbers and a < b. Then a 2 b 2. From
consequence 4 above it follows that there is a rational number c so that a 2 c b 2. But then
c
a
b.
2
Consequence 5.2.6. There is an irrational number between any two distinct rational numbers.
So the rational numbers and the irrational numbers are densely interspersed throughout our
number system.
71
Copyright: Gary Harris, 2009
Exercises for Section 5.2.
Exercise 5.2.1. Let L a 1 such that a 2 2 . Show that is in L, but is not an upper
bound for L. (Find something in L that is bigger than ) Next show that is an upper bound for
L but is not the least upper bound for L. (Find an upper bound for L that is smaller than )
Exercise 5.2.2. Suppose that a and b are two rational numbers. Show that the average of a and
b is also a rational number.
Exercise 5.2.3. Suppose a and b are any two distinct rational numbers. Show that there is a
rational number r so that a < r < b.
Exercise 5.2.4. Let a be a rational number show that a 2 is not a rational number.
Exercise 5.2.5. Let a be a rational number show that a 2 is not a rational number.
Exercise 5.2.6. Suppose that r and s are both irrational numbers. Is it possible for rs
(r times s) to be a rational number?
Explain.
Exercise 5.2.7. Show there is no biggest rational number.
Exercise 5.2.8. Show there is no biggest irrational number.
Exercise 5.1.9. Show there is no smallest positive rational number.
Exercise 5.1.10. Show there is not smallest positive irrational number.
72
Copyright: Gary Harris, 2009
Section 5.3
Quadratic Equations
Now that we know about square roots we can think about solving equations that include squares.
Consider the equation
We wish to find values for x that satisfy this equation. We apply equal actions, additive inverse,
and additive identity to get the equation
The idea is to add the same number to both sides so that the left side of the equation becomes a
perfect square. We want to fill in the two blanks in the following expression:
( ))
( )
(
Let’s recall how we square a sum with two terms:
(
)
(
)(
)
We observe that the expression on the right side of this equation is a perfect square exactly when
the constant term is the square of half the number in front of x. In the problem in which we are
trying to make a perfect square, the number in front of x is 5. So what we need to put in the right
blank space is ( ) , and the number that goes in the left blank space is . It is easy to check that
(
( ))
( )
So back to our original problem: Find the values of x so that
We add ( ) to both sides to get
( )
( )
The left side is a perfect square. We do some arithmetic on the right side and obtain
( )
(
)
Thus
(
√
With equal actions and simplification we obtain
( )
√
)
√
The process we have just completed is called completing the square.
73
√
Copyright: Gary Harris, 2009
We now consider the general quadratic equation:
We multiply both sides by the multiplicative inverse of a and then proceed to mimic the steps we
did above:
(
(
)
(
)
)
(
(
)
)
√
√
√
We have just derived the well-known quadratic formula. There is still one small problem: If
we cannot find a rational or irrational number that works for x. (See Exercise
5.3.1.) Even with all our efforts to construct the “real number” system, we still cannot solve a
simple quadratic equation like
Such is life, or at least, such is mathematics.
Exercises for Section 5.3.
Exercise 5.3.1. Prove that there does not exist a rational or irrational number whose square is
negative.
74
