capstone mathematics for future teacherszimmer.csufresno.edu/~lburger/math 149 book.pdf · 1.3.1 at...

$: Capstone Mathematics for Future Teacherszimmer.csufresno.edu/~lburger/Math 149 Book.pdf · 1.3.1 At the Primary level - The Proportion Area Model The topic of continued fractions$
Capstone Mathematics for Future Teachers

Lance Burger

Fresno State Preliminary Edition

Contents

Preface ix

1 Introduction to Q−Fractions 11.1 Two Definitions of the Rational Numbers . . . . . . . . . . . . 1

1.2 Unit Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . 71.3.1 At the Primary level - The Proportion Area Model . . 71.3.2 Reversibility and Teaching Mathematics . . . . . . . . 10

1.4 Chapter 1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Z− The Integers 152.1 What are the Integers?-Groups . . . . . . . . . . . . . . . . . 15

2.2 The Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . . 17

2.2.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2.2 Division Algorithm . . . . . . . . . . . . . . . . . . . . 172.2.3 Proof of the Euclidean Algorithm . . . . . . . . . . . . 18

2.3 Linear Diophantine Equations in Z . . . . . . . . . . . . . . . 222.3.1 Solving ax+ by = c . . . . . . . . . . . . . . . . . . . . 222.3.2 Linear Diophantine Equation Proofs . . . . . . . . . . 23

2.4 What is the best the Integers can Be?-Rings . . . . . . . . . . 272.5 Chapter 1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 30

v

vi CONTENTS

3 Area Models in Pedagogy 333.1 Fraction Addition and Subtraction . . . . . . . . . . . . . . . 33

3.1.1 Pattern Blocks . . . . . . . . . . . . . . . . . . . . . . 333.2 Fraction Multiplication . . . . . . . . . . . . . . . . . . . . . . 353.3 Fraction Division . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.3.1 Measurement and Partitive Verbal Storylines . . . . . . 363.4 Algebra Tiles and the Quadratic Formula . . . . . . . . . . . . 383.5 Chapter 3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Project-Based Learning 434.1 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2 Chapter 4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 44

Preface

Lance Burger

ix

Chapter 1

Introduction to Q−Fractions

One theme of this book to teach prospective mathematics teachers how todesign lessons less vertically, and more horizontally. What is meant by this isthat most mathematics instruction begins with the basics of a topic and thenbuilds on the basics so that they can be applied in what are typically knownas word problems. These types of problems most often occur at the endof a Chapter or section of a text. One problem with this approach is thatwhen building blocks are introduced without much context or connectionwith each other, students often can not see their relevance or use, and thusquite logically ’tune out’on the subject. The central theme of this bookthen, is that it is better to practice doing ’many’complex problems whichrequire a lot of sub-procedures, concepts and problem solving strategies, thanit is to learn and drill the basics first, with less treatment on problem solvinglater. In light of recent ’Common Core’trends in mathematics which focus onproblem solving, this more complex horizontal/holistic repetition approachhopefully can help students as they will hopefully need less memorizationof procedures which have been learned by working with richer conceptualstructures.

1.1 Two Definitions of the Rational Numbers

The origins of continued fractions are diffi cult to pinpoint exactly, but histor-ical records date their use to at least 2000 years ago. In simplified language,continued fractions historically represent numerical quantities by having onlya 1 in the numerator of a fraction. Just some basics, fractions are elements

1

2 CHAPTER 1 INTRODUCTION TO Q−FRACTIONS

of the Rational numbers, denoted by the script symbol Q, we have two im-portant definitions:

1. Q = {pq: p and q are integers, and q 6= 0}.

2. Rational numbers are numbers expressible as repeating decimals.

• What are the integers? Do you know the symbol for them?

• Why must the definition for the rational numbers have b 6= 0? Couldyou explain why if a student asked?

• What are the names for the different parts of the fraction, p and q inthe first definition above?

• How does the number 5.2 fit in with the second definition for rationalnumbers?

• Is 2.45in the form of a rational number, by the definitions above?

Lecture Question 1.1 Is 0.999 . . . < 1?

Lecture Question 1.2 What fraction is 1.24682468 . . .?

Lecture Question 1.3 What fraction is 23.1452378?

Example 1.1 What the 2, 000, 007th digit after the decimal point for theprevious problem?

Since the repeating block has length of 4 digits, we need to know howmany times 4 goes into 2000007 − 3 = 2000 004. Three is subtracted sincethere are three digits, 145 after the decimal place before the repeating blocksbegin, so those digits are part of the 2, 000, 007 digits after the decimal andneed to be counted. Thus,

2, 000, 004

4= 500, 001 with no remainder, so the answer is 8.

Remark 1.1 If the remainder was 1, the answer would be 2, if the remainderwas 2 then 3 would be the answer. A remainder of 4 cannot happen whendividing by 4. Why?

1.2 UNIT FRACTIONS 3

1.2 Unit Fractions

In the distant past, it did not make much sense to people to divide a smallernumber by a larger one - which is one reason continued fractions were ap-pealing. Also, peoples such as the Egyptian scribes, preferred to representfractions as sums of ’unit fractions,’which were written as the sum of pro-gressively DECREASING fractions . One reason for this, which is also anexcellent pedagogy tool for primary students, is that different fractions canbe compared, as in the following example.

Example 1.2 Which is larger, 4/5 or 7/10?

One way to compare the fractions is to cross multiply, but this can be adiffi cult process to understand.

4

5?7

10→ 4 · 10 > 5 · 7→ 40 > 35

?→ 4

5>7

10

Another way to compare fractions is to convert them to the same denom-inator. In this case, 10 would suffi ce as a suitable denominator>

4

5>7

10because

2

2· 45=8

10>7

10

But, instead of using the lcm(5, 10) = 10, one could use the common mul-tiple, 5 · 10 = 50, which is not necessarily the smartest common denominatorto use, but it explains the process:

10

10· 45?7

10· 55→ 40

50>35

50→ 50 · 40

50> 50 · 35

50→ 40 > 35

Remark 1.2 Understanding WHY processes and formulas in mathematicswork is an important component to encouraging a culture of instruction thatteaches students to take time to think, reflect and solve problems; since, ifwe want people to create NEW things in mathematics and science, they needto learn how to analyze and understand current knowledge, be it abstract orconcrete.


As Egyptian fractions, we see much more concretely that 45is larger than

710by exactly 1

10:

4

5=1

2+1

5+1

107

10=1

2+1

5

Example 1.3 Egyptian fraction decompositions are NOT unique:

1 =1

2+1

3+1

6

3

4=1

2+1

4; but from above, one can multiply by

1

4to get:

1

4· 1 = 1

4· 12+1

4· 13+1

4· 16=1

8+1

12+1

24, so:

3

4=1

2+1

8+1

12+1

24demonstrating that egyptian fraction representations can actually be writtenin an infinite number of ways.In this section, we will focus on two of these methods for producing

Egyptian fraction representations:

1. Sylvestor’s Method (Also known as the ’Greedy Algorithm’)

2. Divisor Decomposition Method

Sylvester’s Method

Originally developed by Fibonacci (1175-1250), Sylvester rediscovered it in1880.Strategy:

• Subtract from the given non-unit fraction the largest unit fraction pos-sible.

• If the result is not a unit fraction, repeat the procedure as many timesas necessary to obtain all unit fractions.

1.2 UNIT FRACTIONS 5

Example 1.4 34: What is the largest unit fraction less than 3/4?

12is less, so subtract:

3

4− 12=3

4− 24=1

4

3

4=1

2+1

4

Example 1.5 1 : What is the largest unit fraction less than 1?

Again, 12is less, so subtract:

1− 12=1

2

1

2− 13=3

6− 26=1

6

1 =1

2+1

3+1

6

Divisor Decomposition Method

Strategy:

• Make the denominator of the original fraction large enough so that thenumerator can be decomposed in such a way as to produce cancella-tions.


Example 1.6 11/15 : the key to this method is to multiply top and bottomby a number such that the new numerator can be decomposed into numberswhich divide the new denominator.

2

2· 1115=22

30

30 has divisors 1, 2, 3, 5, 6, 10, 15

22 = 15 + 6 + 1

11

15=22

30=15 + 6 + 1

30=15

30+6

30+1

30=1

2+1

5+1

30

Example 1.7 13/24 :

3

3· 1324=39

72

39 = 36 + 3

13

24=39

72=36 + 3

72=36

72+3

72=1

2+1

24

Lecture Question 1.4 Use the Divisor Decomposition Method to write athree-term unit fraction decomposition of 7

8.

Lecture Question 1.5 Use the Divisor Decomposition Method to write athree-term unit fraction decomposition of 4

11.

1.3 CONTINUED FRACTIONS 7

1.3 Continued Fractions

Definition 1.1 An expression of the form a0+ 1a1+

1

a2+1

a3+···

is a simple con-

tinued fraction where the ai can be either real or complex numbers, howeverfor this text, they will be taken as positive integers. As a short-hand, thecontinued fraction defined can be written as: [a0, a1, a2, a3, ...]. If the fractiondoes not contain a ’whole number’part, then a0 = 0.

• What is another name and the symbol for the positive integers?

• If the number 0 is added to the set of positive integers, what is thename and symbol for this new set?

• What if a student asked, why does 132

= 23? Could you provide a math-

ematical justification?

Example 1.8 The fraction 45= [0, 1, 4]... since: 4

5= 0 + 1

54

= 0 + 11+ 1

4

.

Example 1.9 The fraction 75= [1, 2, 2]... since upon using the division al-

gorithm two times: 75= 1 + 2

5= 1 + 1

52

= 1 + 12+ 1

2

.

1.3.1 At the Primary level - The Proportion AreaModel

The topic of continued fractions is a rich topic spanning from the primarygrades all the way to the highest levels of mathematical thought. For exam-ple, from the Common Core standards for 3rd grade:

(4) Students describe, analyze, and compare properties of twodimensional shapes. They compare and classify shapes by theirsides and angles, and connect these with definitions of shapes.Students also relate their fraction work to geometry by expressingthe area of part of a shape as a unit fraction of the whole.

The Proportion Area Model is based on the idea of modeling a lwas a

rectangle of length l and width w. The area concept for this model onlyrelates to ’decomposing’the rectangular area into the maximal number of


squares having proportion 1. When a largest number of squares has beenmade, the learner continues to make the next largest size. This processgenerates the continued fraction.

Example 1.10 Using the proportion area model, find the continued fractionof 45

16.

4516is first represented as a rectangle having he following proportions:

Figure 1

Next, by simply making the largest possible square within the rectangle,Figure 2 is obtained:

Figure 2 The division algorithm using squares.

This process of making squares in the rectangle, if you think about it, is thesame as seeing how many times 16 divides into 45, which is twice withremainder 13. Notice that the two 16× 16 squares are each in a proportionof 1 and the remaining rectangle is in a proportion of 13

16, which corresponds

to: 4516= 2 + 13

16. To ’continue’to use the division algorithm until the last


fraction is a unit fraction, it is necessary to change 1316into 1

1613

, which is kind

of like now turning one’s head sideways and doing the previous ’largestsquare’process on the orange 13× 16 rectangle.

Figure 3

The representation in Figure 3 corresponds to: 4516= 2+ 1

1+ 313

, so since 313

is not a unit fraction, the process must continue on the red 3× 13 rectangle,resulting in Figure 4:

Figure 4 The final diagram for the continuedfraction of 45/16.

.

Since the last yellow rectangle is 1 × 3 the process MUST stop, since 131

would keep resulting in the same answer of 13, and we see the correspondence

of counting the largest to the smallest squares to the representation:

4516= 2 + 1

1+ 1

4+13

= [2, 1, 4, 3].


Lecture Question 1.6 Use the Proportion Area Model to make a continuedfraction rectangle for 49

16.

1.3.2 Reversibility and Teaching Mathematics

In Piaget’s theory of Cognitive Development, the Concrete Operational Stagerefers to the ability reason logically on abstract objects such as numbers, frac-tions and variables. A process that aides a learner to develop this capabilityis known as ’Reversibility,’which is the ability to recognize that numbers,objects, or expressions can be changed and then returned to their originalconditions. As an example for how to model instruction using reversibility,after making the continued fraction diagrams in the previous section, stu-dents can be given a final diagram, such as in Figure 4, and work backwardsto find the original fraction.

Example 1.11 Find the original fraction which resulted in the continuedfraction diagram in Figure 5 below:

Figure 5 Working backwards.

This diagram reads as: [2, 5, 2] = 2 + 15+ 1

2

. Working from the bottom,

5+ 12= 2

2· 51+ 1

2= 10

2+ 1

2= 11

2; thus, [2, 5, 2] = 2+ 1

112

= 2+ 211= 11

11· 21+ 2

11=

2211+ 2

11= 24

11.

As seen in the previous example, many aspects of fraction arithmetic wereneeded in order to solve the problem. This idea is a common theme of thisbook, which is to teach rich problems which involve many standards and


procedures, as opposed to having your future students working too much onthe basics in isolation why those basics are relevant.

Lecture Question 1.7 What fraction is [1, 2, 3, 4, 5, 6]?

Algebraic Connections-Finding Square Roots

Example 1.12 Prove√2 = [1, 2, 2, 2, 2, ...].

Proof. Let x = 1+ 12+ 1

2+···then the key is to look at the continued fraction

and decompose a number so that the entire fraction is again seen, as in thefollowing:

x = 1 +1

2 + 12+···

= 1 +1

1 + 1 + 12+···

= 1 +1

1 + x

x = 1 +1

1 + x→ x(1 + x) = (1 + x) · 1 + 1

x2 + x = x+ 1 + 1 = x+ 2

x2 + x = x+ 2→ x2 = 2→ x = ±√2

But, since the continued fraction is obviously positive, then it has been shownthat x =

√2

Lecture Question 1.8 Prove that√40 = [6, 3, 12].


1.4 Chapter 1 Exercises

Exercise 1 Use the Proportion Area Model and graph paper to make acolored continued fraction rectangle for 29

17.

Exercise 2 Use the Proportion Area Model and graph paper to make a col-ored continued fraction rectangle for 26

19.

Exercise 3 Write 37as a a sum of unit fractions using Sylvester’s method.

Exercise 4 What fraction is [3, 4, 1, 1, 2, 5]?

Exercise 5 Write 1118as a sum of unit fractions using the Divisor Decompo-

sition Method.

Exercise 6 Without graph paper, use the division algorithm to sketch a Pro-portion Area Model for 1885

102.

Exercise 7 Recall that the Fibonacci sequence is generated by letting F1 = 1,F2 = 1 and Fn = Fn−1 + Fn−2.

1. Generate the first 20 terms of the Fibonacci sequence.

2. Convert successive ratios of Finonacci numbers into continued fractions.In other words, given 1/1, 2/1, 3/2, 5/3, 8/5, . . . .find the continuedfractions for the ratios listed. Describe the pattern.

Exercise 8 Prove√5 = [2, 4, 4, 4...].

Exercise 9 Prove√7 = [2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4...].

1.4 CHAPTER 1 EXERCISES 13

Exercise 10 Write a 1-2 page reflection on the ’Polished Stones’ video.What aspects of mathematics instruction in Taiwan and Japan do you likeor dislike? In what ways can instruction in the US be informed by whatyou saw? In what ways could instruction in Taiwan and Japan benefit bymethods used in the US?

Exercise 11 Write the repeating decimal, 23.00084538453... as a fraction.Show your solution method.

Exercise 12 Given the fraction 11324499,990,000

, determine the 5000th digit afterthe decimal point of its decimal representation. Show your solution method.

Chapter 2

Z− The Integers

2.1 What are the Integers?-Groups

Most all of you can recall that integers are the counting numbers, along withtheir opposites. One of the integers, though, is its own opposite, namely 0needs to be included, thus:

Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .}

More precisely in terms of what is called ’abstract algebra,’the integersare a GROUP.

Definition 2.1 A Group (G, ∗) is a set with a binary operation ∗, havingthe following properties:

1. CLOSED-The set G is ’closed’under the operation ∗. In the integersthe operation is +, or addition, and adding any two integers is still aninteger, so everything stays in the set. For example, 309+(−26) = 283and 283 is an integer.

2. ASSOCIATIVE-The operation * is associative, meaning: (a ∗ b) ∗c = a ∗ (b ∗ c). For example, (2 + 3) + (−5) = 5 + (−5) = 0 and2 + (3 + (−5)) = 2 + (−2) = 0.

3. IDENTITY- The set G has an identity element, e, for the operation ∗,such that: e ∗ a = a ∗ e = a for every element a in G. (a ∈ G). Forthe integers e = 0 and 0 + a = a+ 0 = a for all integers.

15

16 CHAPTER 2 Z− THE INTEGERS

4. INVERSES-For every element a in the set G there is an inverse elementb ∈ G such that a ∗ b = b ∗ a = e. Typically, for an additive groupb = −a and for a multiplicative group b = a−1. As an example in theintegers, the inverse of the element 234 is −234 since 234 + (−234) =(−234) + 234 = 0.

Definition 2.2 A binary operation ∗ is COMMUTATIVE if a ∗ b = b ∗ afor ALL of the elements in the set the operation is defined on.

Remark 2.1 It is important to point out that in the definition for a group,commutativity is required for identity elements and inverse elements, but thisdoesn’t mean the operation is commutative for ALL of the elements in thegroup.

Definition 2.3 An ABELIAN GROUP ,named after the mathematician NielsAbel (1802-1829), is a group (G, ∗) such that a ∗ b = b ∗ a for ALL of theelements in G.

• Is the set of natural numbers {1, 2, 3, . . .}, closed under the operationof subtraction?

• Are the integers closed under the operation of division?

• Is the set of integers a group under the operation of multiplication?

• Is subtraction an associative operation in the integers? Can you givea counter example?

• Is (Z,−) a group?

• Is (Z, ·) a group?

• Is (Z,÷) a group?

• Is (Z,+) an abelian group?

• Is the subtraction operation commutative for the integers? Note thatsubtraction is defined as: a− b = a+ (−b).

2.2 THE EUCLIDEAN ALGORITHM 17

2.2 The Euclidean Algorithm

2.2.1 History

Though the algorithm was coined after mathematician Euclid after appear-ing in Euclid’s Elements around 300 BC, the algorithm itself is thought tohave been in existence since around 500 BC. Euclid’s method however, wasapplied geometrically as a method to find a common measure between twoline lengths.

2.2.2 Division Algorithm

Before going over the Euclidean Algorithm, I will first describe the DivisionAlgorithm, the backbone of the Euclidean Algorithm.

Definition 2.4 Division Algorithm. Given any integer a and b, with a > 0,there exists a unique integer q and r such that b = qa+ r, where 0 ≤ r < a.The integer a is called the DIVIDEND, b is called the DIVISOR, q is calledthe QUOTIENT, and r is called the REMAINDER.

Example 2.1 To show the simplicity in the Euclidean Algorithm, we go overa short example.

Let b = 325 and a = 53. Solve for q and r.We begin the example by dividing the smaller number, a, into the larger

number, b to get 325/53 = 6.1320. We can easily see that 6 is the quotient.To find the remainder, we go through a few short steps. We multiply thequotient by the smaller number to get, 53x6 = 318. Then, we subtract thatresult from the larger to number to get, 325 − 318 = 7. Therefore, we findthat 7 is our remainder, and in all, q = 6 and r = 7.Now, the Euclidean Algorithm is simply an application of the Division

Algorithm.


2.2.3 Proof of the Euclidean Algorithm

Proof. Given integers a and b with b > 0, we make repeated applications ofthe Division Algorithm to obtain a series of equations:

a = bq1 + r1,

where 0 < r1 < b. This continues with,

b = r1q2 + r2,

r1 = r2q3 + r3,

with 0 < r2 < r1 and 0 < r3 < r2 until we reach,

rj − 1 = rjaj + 1

where gcd(a, b) = rj, the last nonzero remainder.Proof. The proof is very simple. The chain of equations is obtained bydividing b into a, r1 into b, r2 into r1, and so forth until rj is divided intorj − 1.The process of division stops when the remainder is 0. A key difference

between the Division Algorithm and the Euclidean Algorithm is the equalitysigns on the conditions.For the Division Algorithm 0 ≤ r < b, however, for the Euclidean Algo-

rithm, it is simply 0 < r < b. The reason being is that if r = 0, then theseries of equations would stop at a = bq1, implying the gcd(a, b) = b.The key idea of the proof is to prove that rj is the greatest common

divisor or a and b. First, we observe the following:

gcd(a, b) = gcd(a− bq1, b)

= gcd(r1, b)

= gcd(r1, b− r1q2)= gcd(r1, r2)

= gcd(r1 − r2q3, r2)= gcd(r3, r2)

By mathematical induction, we end up seeing that gcd(a, b) = gcd(rj −1, rj) = gcd(rj, 0) = rj. Concluding the proof that rj = gcd(a, b).


Corollary 2.1 As a consequence of the above proof, back substitution andaccumulation of terms can be performed, similar to the following example, sothat a linear combination can always be found such that:

ax+ by = gcd(a, b),

which is the key to solving linear diophantine equations!

Example 2.2 Find the greatest common divisor of 42823 and 6409.

We first let b = 42823 and c = 6409. By applying the Division Algorithm,we divide c into b to see that 42823/6409 = 6.6816976. Making the quotientq1 = 6. Then, we multiply the quotient by c to get 6 × 6409 = 38454.Finally, we subtract 38454 from b to get 42823− 38454 = 4369, making 4369the remainder, r1.Next, we apply the Euclidean Algorithm to see the series of equations:

42823 = 6× 6409 + 4369

6409 = 1× 4369 + 2040

4369 = 2× 2040 + 289

2040 = 7× 289 + 17

289 = 17× 17 + 0Because the last remainder of the algorithm is 0, the gcd(42823,6409) =

17.

Definition 2.5 The integers have the DISTRIBUTIVE PROPERTY formultiplication OVER addition, or a · (b+ c) = a · b+ a · c for all elements a, band c in Z.

Definition 2.6 In the integers, we say that n | a if there is an integer b suchthat a = n · b. For exampe, 6 | −24 because there is an integer, −4, suchthat 6 · (−4) = −24. Verbally, n | a is said as n ’divides’a.


Summary Reflections 1 Based on the previous definition does 0 | 12 makeany sense?

Definition 2.7 The gcd(a, b), or greatest common divisor of two integers aand b, is the largest positive integer that divides both a and b.

Definition 2.8 A PRIME number is a positive integer p with exactly twofactors, p and 1.

Example 2.3 How many positive factors are there for the integer 75?

75 = 31 · 52

The possible factors are all of the combinations of {30, 31} × {50, 51, 52} .There are 2× 3 = 6 possible factors.

1. 30 · 50 = 1 · 1 = 1

2. 30 · 51 = 1 · 5 = 5

3. 30 · 52 = 1 · 25 = 25

4. 31 · 50 = 3 · 1 = 3

5. 31 · 51 = 3 · 5 = 15

6. 31 · 52 = 3 · 25 = 75

As a quick way to solve problems such as this, simply write the primefactoriation and add 1 to each of the powers to account for the 0 power pos-sibility that none of that prime is a factor, and then multiply the (powers+1)together:

31 · 52 → 31+152+1 → 2 · 3 = 6.

Lecture Question 2.1 How many positive factors does 5400 have?

Definition 2.9 Integers a and b are RELATIVELY PRIME iff gcd(a, b) =1.


Definition 2.10 n is an EVEN integer iff 2 | n , or n = 2k, for some k ∈ Z.If n is not even, then it is ODD and of the form n = 2k+1 for some k ∈ Z.

Example 2.4 Reduce the fraction 4256by finding the gcd(42, 56).

One way to find the gcd of a group of integers is to find the common primefactors and multiply them together. 42 = 6 ·7 = 2 ·3 ·7 and 56 = 8 ·7 = 23 ·7so the common prime factors are 2 and 7, which multiply to 14. 42÷ 14 = 3and 56÷ 14 = 4. This means that 42

56= 14

14· 34= 1 · 3

4= 3

4.

Example 2.5 The ’Calculator Trick’ is a quick way to find gcd′ s and isdemonstrated here as a way to reduce a diffi cult fraction using the EuclideanAlgorithm:

Reduce the fraction 80518633

.The following use of the calculator works because of the distributive prop-

erty.

1. 8633÷ 8051 = 1.072289157... (Since this is a rational number, it mustbe a repeating decimal.)

2. Now, subtract offthe ’integer part’of the decimal number: 1.072289157...−1 = 0.072289157...

3. Next multiply this decimal part by the original divisor: 8051·(0.072289157) =582 This is the remainder. Again, why this works is because of thedistributive property. 8633

8051= (1 + 0.072289157...). After cross multi-

plying the 8051 one obtains: 8633 = 8051 ·1+8051 · (0.072289157...) =8051 + 582 Since the decimal part is less than 1, it must be that notall of an 8051 went into 8633, which is why it is the remainder. TheEuclidean algorithm now continues by dividing 8051 by 582.

4. 8051÷582 = 13. 8333... Again, using the calculator trick, 13.833−13 =0.833 and 582 · (0.83333...) = 485.

5. 582÷ 485 = 1. 2. 1.2− 1 = 0.2. 485 · (0.2) = 97.

6. 485÷ 97 = 5 This is the final step of the Euclidean Algorithm, sincethere is no remainder. This imlplies that gcd(8051, 8633) = 97.

7. Now it is much easier to reduce the fraction, since we know the largestcommon factor of both numbers. 8051÷ 97 = 83 and 8633÷ 97 = 89.The answer is 83

89.


2.3 Linear Diophantine Equations in Z

2.3.1 Solving ax+ by = c

Definition 2.11 A DIOPHANTINE EQUATION is any equation in whichthe solutions are ’over’the integers.

Remark 2.2 The most famous diophantine equation is xn+ yn = zn, whichis known as ’Fermat’s Last Theorem, which was conjectured by the French-man Pierre de Fermat in 1637 and finally proven by the British mathemati-cian Andrew Wiles in 1995. The theorem ’conjectured’that for n > 2, thereare no integer solutions.

Figure 1Diophantus.

Example 2.6 x3+y3 = z3 has many solutions over the real numbers R, such

as x = 1.y = 1 and z = 3√2; but it is known now that there are no solutions

over the integers, due to Fermat’s Last Theorem.

The termDiophantine derives from the ancient Greek mathematician Dio-phantus, known as the ’Father of Algebra.’ Diophantus lived in Alexandriaaround AD 250 and wrote one of the earliest books on algebra, Arithmetica.

• The equation 2x+ 3 = 4 is not a Diophantine equation. Why not?

• The equation 2x + 6y = 12 is a Diophantine equation. Can you findseveral solutions in integers? How many solutions are there?

• The simplest Diophantine equation in a single variable is ax = b, witha 6= 0. Under what condition is it solvable in integers?

2.3 LINEAR DIOPHANTINE EQUATIONS IN Z 23

2.3.2 Linear Diophantine Equation Proofs

Preliminaries:

Definition 2.12 The integers have the DISTRIBUTIVE PROPERTY formultiplication OVER addition, or a · (b+ c) = a · b+ a · c for all elements a, band c in Z.

Definition 2.13 In the integers, we say that n | a if there is an integer bsuch that a = n · b. For exampe, 6 | −24 because there is an integer, −4,such that 6 · (−4) = −24.

Lemma 1 Given a, b, c, d, n ∈ Z, if n | a and n | b then n | ac+ bd.

Proof. Since n | a and n | b, then there must be integers k and l such thata = nk and b = nl. So, upon substitution: ac+ bd = nkc+ nld = n · (kc+ld). Since the integers are CLOSED under addition and multiplication, thenkc+ ld ∈ Z, so this implies that n | ac+ bd

Remark 2.3 The reverse of the DISTRIBUTIVE PROPERTY was usedabove to obtain nkc+ nld = n · (kc+ ld). This is called FACTORING.

Here is the main Theorem for solving Diophantine Equations in this Chap-ter.

Theorem 1 Let a, b, c ∈ Z. Consider the Diophantine equation

ax+ by = c.

Claim 1: If gcd(a, b) - c then there are no solutions.Claim 2: If gcd(a, b) | c, then there are infinitely many solutions of the

form:

x = x0 +b

gcd(a, b)· k

y = y0 −a

gcd(a, b)· k

where (x0, y0) is a ’particular solution,’and k ∈ Z.


Claim 3: Any integer solution (x, y) to ax+ by = c has the form in part(b), given ax0 + by0 = c.

Proof. Claim 1-(Proof by contradiction) Suppose gcd(a, b) - c and thereIS a solution (x, y) . Then by the previous Lemma, since gcd(a, b) | a andgcd(a, b) | b, then gcd(a, b) | ax+ by = c→← (these two implication arrowsgoing opposite directions means there is a contradiction in the logic, so itmust be the case that if gcd(a, b) - c then there are NO solutions.)Claim 2-Suppose gcd(a, b) | c. Then by definition, there is an integer k

such that c = k ·gcd(a, b). By the previous theorem, there are integers s andt such that:

as+ bt = gcd(a, b)

therefore;ask + btk = k · gcd(a, b) = c.

giving a ’particular solution;’

x0 = sk

y0 = tk.

Given this particular solution, the general solution now follows:

a

(x0 +

b

gcd(a, b)· k)+ b

(y0 −

a

gcd(a, b)· k)

= ax0 +abk

gcd(a, b)+ by0 −

bak

gcd(a, b)

= ax0 + by0 +abk

gcd(a, b)− abk

gcd(a, b)

= ax0 + by0 = c

Claim 3-Suppose ax+ by = ax0 + by0 = c then,

ax+ by − ax0 − by0 = c− c = 0= ax− ax0 + by − by0= a(x− x0) + b(y − y0)

a(x− x0) = b(y0 − y)

But recall that d = gcd(a.b), by definition, divides a and b,and since it is thegreatest common divisor, it makes sense that gcd(a

d, bd) = 1 (Euclid’s lemma).

2.3 LINEAR DIOPHANTINE EQUATIONS IN Z 25

a

d· (x− x0) =

b

d· (y0 − y)

gcd(a

d,b

d) = 1→ a

d| (y0 − y)→ y0 − y =

a

d· k , for some k ∈ Z

a

d· (x− x0) =

b

d· ad· k → (x− x0) =

b

d· k

x = x0 +b

d· k

y = y0 −a

d· k

Example 2.7 Solve the Diophantine equation 42823x+ 6409y = 51.

The first step in solving a Diophantine equation, is to check to see if it iseven solvable. To do that, it is necessary to use the previous listed theoremsand find the gcd(a, b) and then see if the gcd divides the c terms. In Example2, it was found that gcd(42823, 6409) = 17. Since 17 | 51 , i.e., 51 = 3 · 17,then this Diophantine equation is solvable. The next step is to find integersx and y such that:

42823x+ 6409y = 17

To do this, a series of substitutions are required. Going back to Example2, recall the following steps in finding the gcd, but this time, in each case,solve for the remainder as follows:

42823 = 6× 6409 + 4369→ 42823(1) + 6409(−6) = 4369

6409 = 1× 4369 + 2040→ 6409(1) + 4369(−1) = 2040

4369 = 2× 2040 + 289→ 4369(1) + 2040(−2) = 289


2040 = 7× 289 + 17→ 2040(1) + 289(−7) = 17

Now, beginning with the last equation, a long string of back substitutionsand distributive multiplication must be performed as follows:

2040(1) + 289(−7) = 17

2040(1) + [4369(1) + 2040(−2)](−7) = 2040(1) + 4369(−7) + 2040(14)= 2040(1 + 14) + 4369(−7)= 2040(15) + 4369(−7) = 17

2040(15) + 4369(−7) = [6409(1) + 4369(−1)](15) + 4369(−7)= 6409(15) + 4369(−15) + 4369(−7)= 6409(15) + 4369(−22) = 17

6409(15) + 4369(−22) = 6409(15) + [42823(1) + 6409(−6)](−22)= 6409(15) + 42823(−22) + 6409(132) = 17

42823(−22) + 6409(147) = 17

Hence, x = −22 and y = 147 for the equation 42823x+6409y = 17. Nowto solve the original problem, since 51 = 3 ·17, we can multiply both sides ofthe gcd equation by 3 and distribute it in to the x and y solutions as follows:

3[42823(−22) + 6409(147)] = 3 · 17

42823(3 · (−22)) + 6409(3 · 147) = 3 · 17

42823(−66) + 6409(441) = 51

The solution to the original problem is: x = −66 and y = 441.

2.4 WHAT IS THE BEST THE INTEGERS CAN BE?-RINGS 27

Definition 2.14 From now on, gcd(a, b) = (a, b).

Example 2.8 Can solutions to the above problem be found in which both arepositive?

Obviously since both 42823 and 6409 are positive and greater than 51,it will not be possible, but another way to see this is to apply the infinitesolution formulae:

x = x0 +b

(a, b)· k

y = y0 −a

(a, b)· k

x = −66 + 640917· k = −66 + 37k

y = 441− 4282317

· k = 441− 2519k

−66 + 37k > 0→ 37k > 66→ k > 0.173...

441− 2519k > 0→ 441 > 2519k → k < 0.175...

Since k must an integer, this wil be impossible, thus we conclude thatboth solutions cannot be positive.

Lecture Question 2.2 If solvable, find all negative integer solutions,

6x+ 15y = −3000.

2.4 What is the best the Integers can Be?-Rings

The ’best’structure (more on this word best later) that can be constructeddirectly from the integers is a ring.


Definition 2.15 A Ring (R,+, •) is an Abelian group (R,+) with a sec-ondary operation · having the following properties:

1. • is CLOSED.

2. • is ASSOCIATIVE, i.e., (a • b) • c = a • (b • c), for all elements in R.

3. DISTRIBUTUVE PROPERTY, i.e., a • (b + c) = a • b + a • c for allelements in R.

Definition 2.16 A ring (R,+, •) is a COMMUTATUVE ring iff · is a com-mutive operation.

Definition 2.17 A ring (R,+, •) is a RING WITH UNITY iff · has anidentity element (often referred to as 1)

Example 2.9 (Z,+, •) is a COMMUTATIVE ring with UNITY.

What do we want the integers to become?- FIELDS!

In this chapter, we were introduced to the ring of integers, which of course,lacks multiplicative inverses. In the first chapter, the rational numbers didhave inverses for all non-zero elements, and comprised our first example ofthe following structure:

Definition 2.18 A Field (F,+, •) is comprised of two Abelian groups, (F,+)and (F \ {0}, •),related by the DISTRIBUTUVE PROPERTY, with the con-dition that 0 6= 1, i.e., the identities are distinct.

Example 2.10 Some commonly known fields:

(Q,+, ·) ⊂ (R,+, ·) ⊂ (C,+, ·).

This book has several major themes:

• Appreciate the connections between the Fundamental Theorems ofArithmetic and Algebra;

• Learn the pedagogical power of rich-problem topics.

2.4 WHAT IS THE BEST THE INTEGERS CAN BE?-RINGS 29

• Understand how fields can be constructed from the rings Z and Z[x].

• Appreciate the importance of primary instruction and arithmetic tothe secondary grades and algebra.

In this respect, the next chapter revisits the rational numbers to go overseveral area models which also can be applied to introductory algebra.



Exercise 13 Which of the following Diophantine equations can not be solved?Explain with mathematical justifications.

a. 6x+ 51y = 22

b. 33x+ 14y = 115

c. 14x+ 35y = 93

Exercise 14 Solve the Diophantine equation 56x+ 72y = 40.

Exercise 15 Solve the Diophantine equation 24x+ 138y = 18.

Exercise 16 Solve the Diophantine equation 85x+ 30y = −215.

Exercise 17 One of the Diophantine equations below is solvable. Find thesolution to the solvable equation:

195x+ 221y = −12

57x+ 123y = −12

Exercise 18 Reduce the fraction 773931937

by finding (7739, 31937).

Exercise 19 Reduce the fraction 37355296496

by finding (96496, 373552).

Historical Applications

Exercise 20 775 AD Alcuin of York- A hundred bushels of grain are distrib-uted among 100 people in such a way that each man receives 3 bushels, eachwoman 2 bushels and each child a half a bushel. How many men, womenand children are there?


Exercise 21 850AD-Mahavira-There were 63 equal piles of plantain fruitput together with seven single plantains. They were divided evenly among23 travelers. How many plantains did each traveler get? [Hint: Consider63x+ 7 = 23y.]

Figure 2

Exercise 22 1372 AD-Yen Kung-We have an unknown number of coinswhich were put on a number of strings in equal amounts (see Fig.2). Ifyou make 77 strings of them you are 50 coins short, but if you make 78strings then you have the exact number of coins. Find the number of coins[Hint: Consider 77x− 50 = 78y.]

Exercise 23 Find all positive integer solutions to:

172x+ 20y = 1000.

Exercise 24 Find a solution to the following Diophantine equation, havingan x-value between 17 and 20 :

56x+ 72y = −16.

Exercise 25 Using a prime factor calculator on the web, such as:

http : //www.calculatorsoup.com/calculators/math/prime− factors.php

determine the number of positive factors for 4, 988, 552, 244.

Chapter 3

Area Models in Pedagogy

3.1 Fraction Addition and Subtraction

3.1.1 Pattern Blocks

Pattern blocks are area-based manipulatives, simililar to the pie-chartconcept, often used for modelling fraction arithmetic in the primary grades.The first issue when introducing and using pattern blocks is to determinetheir value. A useful way to introduce pattern blocks is to assign varieties

33

34 CHAPTER 3 AREA MODELS IN PEDAGOGY

of pieces to be ’the whole,’and then have students figure out the values ofthe other pieces. This activity, in its own right, could involve quite a degreeof thinking, and it is well worth devoting adequate time to when introducingthis manipulative.

Remark 3.1 Generally when using pattern blocks for fractions, ther orange-square and tan-narrow-rhombus pieces are not used. These pieces are forinvestigating tilings, as their angles and shapes are needed with the otherpieces in order to tile the plane.

Remark 3.2 Another feature of early introduction to pattern blocks in theprimary grades is to simultaneously introduce basic geometric concepts suchas polygons, regular polygons, hexagons, rhombii, squares, trapezoids, etc..

Lecture Question 3.1 What are the values of the pieces if 1 whole equalsone blue rhombus?

Lecture Question 3.2 What are the values of the pieces if 1 whole equalsone red trapezoid?

To perform addition or multiplication, the basic idea is to convert thepieces into the same time of piece, then combine or subtract out the pieces,and then make the pieces as large as possible, in terms of ’reducing’ thefraction. For these problems, we will focus on ’mixed fractions’ so makeas many wholes first, and then make the largest same color pieces for theremaining blocks.

Lecture Question 3.3 Given that two hexagons is a whole, draw diagramsto model the solution to:

5

6− 13.

3.2 FRACTION MULTIPLICATION 35

Lecture Question 3.4 Given that three hexagons is a whole, draw patternblock diagrams to model the solution to:

5

6− 13.

3.2 Fraction Multiplication

Using pattern blocks is an excellent way to model multiplication of fractionsand help justify to students why the following formula works:

a

b· cd=a · cb · d.

Here is the logic behind the model for a problem such as,

1

2× 13=1

6.

The idea in words is to ’take one of the two pieces of one third.’ So givena pattern block representation of 1

3, one splits it up into two pieces or groups

of pieces, and then takes one of them.The following example visually illustrates the model.

Example 3.1 Given that two hexagons = 1, compute 52× 3

2using pattern

blocks.

1. First draw the RHS fraction (by convention).

2. Next, break the RHS fraction into the number of pieces (or groups ofpieces) as the denominator of the LHS fraction, in this case two equalgroups.


3. The ’numerator’drawing now takes five of the groups that 32was made

into.

4. The result is now reduced to a mixed fraction by first making as manywholes as possible, and then the largest number of same-color pieces.

Lecture Question 3.5 On Friday night, John ate pizza for dinner and had4/5 of the pizza left over. On Saturday, he ate 2/3 of what was left. Usepattern block drawings to figure out how much pizza John had left on Sunday?(Let 5 hexagons = 1 whole)

3.3 Fraction Division

3.3.1 Measurement and Partitive Verbal Storylines

Measurement Model

The measurement model is a verbal way of expressing a division such as

11

2÷ 34

as a statement like: How many 3/4 miles are in 112miles?

Notice that for the measurement model, the numerator and denominatorare the same types of objects (miles in this case) and the answer (or quotient)is a unitless number.

Example 3.2 Write a measurement model storyline for the problem:

31

4÷ 23.

Solution: Suppose that a crate of apples weighs 23pounds. How many

crates of apples does it take to weigh 314pounds?

3.3 FRACTION DIVISION 37

Partitive Model

The partitive model is another way to express division, but quite a bit dif-ferent. For the partitive model, the numerator (dividend) and (denominator(divisor) are different objects. The divisor is just a number, and the quotient(answer) is the same type of object as the numerator.

Example 3.3 Write a partitive model storyline for the problem:

31

4÷ 23.

Solution: Using the same theme of crates of apples, it could be phrasedas: 2

3of some crates of apples weighs 31

4pounds. How much did the entire

amount of crates weigh?


3.4 Algebra Tiles and the Quadratic Formula



For problem #’s 26-30, 1 whole = 4 yellow hexagons.

Exercise 26 Make pattern block drawings to compute 54− 2

8.

Exercise 27 Make pattern block drawings to compute 32×34.

Exercise 28 Make pattern block drawins to compute 38÷ 1

3.

Exercise 29 Use pattern block drawings for the following word problem: Thedolphins at the Newport News Aquarium are fed 7/8 of a bucket of fish eachday. The sea otters are fed 2/3 as much fish as the dolphins. How manybuckets of fish are the sea otters fed each day?

Exercise 30 Write a partitive model storyline for the problem 234÷ 7

8and

solve it with pattern block diagrams.

For problem #’s 31-34, 1 whole = 3 yellow hexagons.

Exercise 31 Aaron operates a hot dog stand. On Wednesday he used 5/2bags of hot dog buns. On Thursday he used 4/3 as many hot dog buns as onWednesday. How many bags of hot dog buns did Aaron use on Thursday?Solve the problem two ways. One way using pattern blocks and the otherway using the area model.

Exercise 32 Compute 1118− 5

6using pattern blocks.

Exercise 33 Compute 23÷ 1

2using a pattern block drawing.


Exercise 34 Write a measurement model storyline for the problem 212÷ 56

and solve it with pattern block diagrams.

Classroom Management Corner

Exercise 35 Draw a 1-panel comic of a humourous classroom managementsituation. The above is an example.

Exercise 36 After viewing the Youtube video ’Unconventional Teacher Tac-tics that Quell Bad Behavior,’write a 1-page minimum word-processed reflec-tion for what you liked or dis-liked about the suggestions, and a few examplesfor how it could relate to your future teaching.

Exercise 37 Use the area model to compute:

3

2× 35.


Exercise 38 Use the area model to compute:

4

5÷ 23.

Exercise 39 Use the area model to compute in base 5:

1315 ÷ 235.


3415 ÷ 135.


3417 ÷ 137.

Exercise 42 Use a colored algebra tile diagram to compute:

(2x− 3y + 1) · (3x− 2y − 1) .

Exercise 43 Use a colored algebra tile diagram to compute:

4x2 − 2y2 − 2xy + 7y + 5x− 6x− y + 2 .

Exercise 44 Divide using a colored algebra tile diagram:

6x2 − 7x− 32x− 3 .


Exercise 45 Solve by ’completing the square’using three colored algebra-tilediagrams:

x2 − 6x = −5.

Exercise 46 Solve by ’completing the square’using three colored algebra-tilediagrams:

x2 + 4x = 12.

Exercise 47 Factor using the ’cloud’method:

132x2 + 41x− 40.

Exercise 48 Factor using the ’cloud’method:

14x2 − 83x− 6.

Exercise 49 Use base x to compute:

(2x2 + 3x− 5

)·(x3 − x− 3

).

Exercise 50 Use base x long-division to compute:

3x7 + 2x6 + x5 − 9x4 − 3x3 − x2 + xx4 − 3x+ 1

Chapter 4

Project-Based Learning

4.1 Scaling

(SA

V

)magnified

=

(1

m

)n(SA

V

)original

.

43

44 CHAPTER 4 PROJECT-BASED LEARNING


Exercise 51 Insert completed ’Scaling Activity Worksheet’into Portfolio.

Exercise 52 Insert completed ’Circle Activity Worksheet’into Portfolio.

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