Deepening Mathematics Instruction: Algebraic Structures for Teachers

Transcription

1 Deepening Mathematics Instruction: Algebraic Structures for Teachers Lance Burger Fresno State Preliminary Edition

2 Contents Preface ix 1 Z The Integers What are the Integers?-Groups The Euclidean Algorithm History Division Algorithm Proof of the Euclidean Algorithm Linear Diophantine Equations in Z Solving ax + by = c Linear Diophantine Equation Proofs What is the best the Integers can Be?-Rings Chapter 1 Exercises Induction and Algebraic Thinking Proof by Mathematical Induction Chapter 2 Exercises Exponents and the Binomial Theorem Set Theory Basics Number Sets Justifications of Exponent Properties p is an Irrational Number v

3 vi CONTENTS 3.5 The Binomial Theorem Some Brief History Proofs Chapter 3 Exercises Exploring Quadratic Equations Parabolas and Interpolation Two Forms of a Quadratic Function The General Form The Vertex or Standard Form Factoring Trinomials The ac-cloud method for factoring ax 2 + bx + c over the Integers Projectile Motion Equations of Motion Chapter 4 Exercises

4 Preface This text is for students who want to become secondary mathematics teachers. With an understanding of this book, it is anticipated that a secondary mathematics teacher will understand the importance of even primary mathematics, in terms of its connection to higher more abstract treatments found at the university level. One aim of the book is to show the structural connections between arithmetic and algebra. With this in mind, students should understand that although they might only teach middle or high school after their university degrees and credentials, this approach aims to teach the deep connections and importance of early arithmetic concepts learned in the primary grades, as they form the building blocks for the algebraic concepts needed as far as the college level, preparing their students one day for Calculus and higher mathematics. Another important role for this workbook, is to introduce the prospective teacher to approaches to lesson design which emphasize the creation of rich problems encompassing many standards, as opposed to a linear build-up from basic standards to more complex problems. In the author s experience, it is better to have students practice solving of a lot of similar hard problems, with repetition, rather than build up to a more limited amount of hard problems, not conducive to the seeing of patterns and generalized knowledge for problem solving strategies. Lance Burger January, 2014 ix

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6 Chapter 1 Z The Integers 1.1 What are the Integers?-Groups Most all of you can recall that integers are the counting numbers, along with their opposites. One of the integers, though, is its own opposite, namely 0 needs to be included, thus: Z = {..., 3, 2, 1, 0, 1, 2, 3,...} More precisely in terms of what is called abstract algebra, the integers are a GROUP. Definition 1.1 A Group (G, ) is a set with a binary operation, having the following properties: 1. CLOSED-The set G is closed under the operation. In the integers the operation is +, or addition, and adding any two integers is still an integer, so everything stays in the set. For example, 309+( 26) = 283 and 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 and 2 + (3 + ( 5)) = 2 + ( 2) = 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). For the integers e = 0 and 0 + a = a + 0 = a for all integers. 1

7 2 CHAPTER 1 Z THE INTEGERS 4. INVERSES-For every element a in the set G there is an inverse element b G such that a b = b a = e. Typically, for an additive group b = a and for a multiplicative group b = a 1. As an example in the integers, the inverse of the element 234 is 234 since ( 234) = ( 234) = 0. Definition 1.2 A binary operation is COMMUTATIVE if a b = b a for ALL of the elements in the set the operation is defined on. Remark 1.1 It is important to point out that in the definition for a group, commutativity is required for identity elements and inverse elements, but this doesn t mean the operation is commutative for ALL of the elements in the group. Definition 1.3 An ABELIAN GROUP,named after the mathematician Niels Abel ( ), is a group (G, ) such that a b = b a for ALL of the elements in G. Is the set of natural numbers {1, 2, 3,...}, closed under the operation of 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? a counter example? Can you give 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? subtraction is defined as: a b = a + ( b). Note that

8 1.2 THE EUCLIDEAN ALGORITHM The Euclidean Algorithm History Though the algorithm was coined after mathematician Euclid after appearing in Euclid s Elements around 300 BC, the algorithm itself is thought to have been in existence since around 500 BC. Euclid s method however, was applied geometrically as a method to find a common measure between two line lengths Division Algorithm Before going over the Euclidean Algorithm, I will first describe the Division Algorithm, the backbone of the Euclidean Algorithm. Definition 1.4 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 called the QUOTIENT, and r is called the REMAINDER. Example 1.1 To show the simplicity in the Euclidean Algorithm, we go over a 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 = We can easily see that 6 is the quotient. To find the remainder, we go through a few short steps. We multiply the quotient by the smaller number to get, 53x6 = 318. Then, we subtract that result from the larger to number to get, = 7. Therefore, we find that 7 is our remainder, and in all, q = 6 and r = 7. Now, the Euclidean Algorithm is simply an application of the Division Algorithm.

9 4 CHAPTER 1 Z THE INTEGERS Proof of the Euclidean Algorithm Proof. Given integers a and b with b > 0, we make repeated applications of the Division Algorithm to obtain a series of equations: a = bq 1 + r 1, where 0 < r 1 < b. This continues with, b = r 1 q 2 + r 2, r 1 = r 2 q 3 + r 3, with 0 < r 2 < r 1 and 0 < r 3 < r 2 until we reach, r j 1 = r j a j + 1 where gcd(a, b) = r j, the last nonzero remainder. Proof. The proof is very simple. The chain of equations is obtained by dividing b into a, r 1 into b, r 2 into r 1, and so forth until r j is divided into r j 1. The process of division stops when the remainder is 0. A key difference between the Division Algorithm and the Euclidean Algorithm is the equality signs on the conditions. For the Division Algorithm 0 r < b, however, for the Euclidean Algorithm, it is simply 0 < r < b. The reason being is that if r = 0, then the series of equations would stop at a = bq 1, implying the gcd(a, b) = b. The key idea of the proof is to prove that r j is the greatest common divisor or a and b. First, we observe the following: gcd(a, b) = gcd(a bq 1, b) = gcd(r 1, b) = gcd(r 1, b r 1 q 2 ) = gcd(r 1, r 2 ) = gcd(r 1 r 2 q 3, r 2 ) = gcd(r 3, r 2 ) By mathematical induction, we end up seeing that gcd(a, b) = gcd(r j 1, r j ) = gcd(r j, 0) = r j. Concluding the proof that r j = gcd(a, b).

10 1.2 THE EUCLIDEAN ALGORITHM 5 Corollary 1.1 As a consequence of the above proof, back substitution and accumulation of terms can be performed, similar to the following example, so that a linear combination can always be found such that: ax + by = gcd(a, b), which is the key to solving linear diophantine equations! Example 1.2 Find the greatest common divisor of and We first let b = and c = By applying the Division Algorithm, we divide c into b to see that 42823/6409 = Making the quotient q 1 = 6. Then, we multiply the quotient by c to get = Finally, we subtract from b to get = 4369, making 4369 the remainder, r 1. Next, we apply the Euclidean Algorithm to see the series of equations: = = = = = Because the last remainder of the algorithm is 0, the gcd(42823,6409) = 17. Definition 1.5 The integers have the DISTRIBUTIVE PROPERTY for multiplication OVER addition, or a (b + c) = a b + a c for all elements a, b and c in Z. Definition 1.6 In the integers, we say that n a if there is an integer b such that a = n b. For exampe, 6 24 because there is an integer, 4, such that 6 ( 4) = 24. Verbally, n a is said as n divides a.

11 6 CHAPTER 1 Z THE INTEGERS Summary Reflections 1 Based on the previous definition does 0 12 make any sense? Definition 1.7 The gcd(a, b), or greatest common divisor of two integers a and b, is the largest positive integer that divides both a and b. Definition 1.8 A PRIME number is a positive integer p with exactly two factors, p and 1. Example 1.3 How many positive factors are there for the integer 75? 75 = The possible factors are all of the combinations of {3 0, 3 1 } {5 0, 5 1, 5 2 }. There are 2 3 = 6 possible factors = 1 1 = = 1 5 = = 1 25 = = 3 1 = = 3 5 = = 3 25 = 75 As a quick way to solve problems such as this, simply write the prime factoriation and add 1 to each of the powers to account for the 0 power possibility that none of that prime is a factor, and then multiply the (powers+1) together: = 6. Lecture Problem 1.1 How many positive factors does 5400 have? Definition 1.9 Integers a and b are RELATIVELY PRIME iff gcd(a, b) = 1.

12 1.2 THE EUCLIDEAN ALGORITHM 7 Definition 1.10 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 1.4 Reduce the fraction 42 by finding the gcd(42, 56). 56 One way to find the gcd of a group of integers is to find the common prime factors and multiply them together. 42 = 6 7 = and 56 = 8 7 = so the common prime factors are 2 and 7, which multiply to = 3 and = 4. This means that 42 = 14 3 = 1 3 = Example 1.5 The Calculator Trick is a quick way to find gcd s and is demonstrated here as a way to reduce a diffi cult fraction using the Euclidean Algorithm: Reduce the fraction The following use of the calculator works because of the distributive property = (Since this is a rational number, it must be a repeating decimal.) 2. Now, subtract off the integer part of the decimal number: = Next multiply this decimal part by the original divisor: 8051 ( ) = 582 This is the remainder. Again, why this works is because of the 8633 distributive property. = ( ). After cross multiplying the 8051 one obtains: 8633 = ( ) = Since the decimal part is less than 1, it must be that not all of an 8051 went into 8633, which is why it is the remainder. The Euclidean algorithm now continues by dividing 8051 by = Again, using the calculator trick, = and 582 ( ) = = = (0.2) = = 5 This is the final step of the Euclidean Algorithm, since there is no remainder. This imlplies that gcd(8051, 8633) = Now it is much easier to reduce the fraction, since we know the largest common factor of both numbers = 83 and = 89. The answer is

13 8 CHAPTER 1 Z THE INTEGERS 1.3 Linear Diophantine Equations in Z Solving ax + by = c Definition 1.11 A DIOPHANTINE EQUATION is any equation in which the solutions are over the integers. Remark 1.2 The most famous diophantine equation is x n + y n = z n, which is known as Fermat s Last Theorem, which was conjectured by the Frenchman Pierre de Fermat in 1637 and finally proven by the British mathematician Andrew Wiles in The theorem conjectured that for n > 2, there are no integer solutions. Figure 1 Diophantus. Example 1.6 x 3 +y 3 = z 3 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 term Diophantine derives from the ancient Greek mathematician Diophantus, known as the Father of Algebra. Diophantus lived in Alexandria around 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 find several solutions in integers? How many solutions are there? The simplest Diophantine equation in a single variable is ax = b, with a 0. Under what condition is it solvable in integers?

14 1.3 LINEAR DIOPHANTINE EQUATIONS IN Z Linear Diophantine Equation Proofs Preliminaries: Definition 1.12 The integers have the DISTRIBUTIVE PROPERTY for multiplication OVER addition, or a (b + c) = a b + a c for all elements a, b and c in Z. Definition 1.13 In the integers, we say that n a if there is an integer b such 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 that a = nk and b = nl. So, upon substitution: ac + bd = nkc + nld = n (kc + ld). Since the integers are CLOSED under addition and multiplication, then kc + ld Z, so this implies that n ac + bd Remark 1.3 The reverse of the DISTRIBUTIVE PROPERTY was used above to obtain nkc + nld = n (kc + ld). This is called FACTORING. Here is the main Theorem for solving Diophantine Equations in this Chapter. 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: b x = x 0 + gcd(a, b) k a y = y 0 gcd(a, b) k where (x 0, y 0 ) is a particular solution, and k Z.

15 10 CHAPTER 1 Z THE INTEGERS Claim 3: Any integer solution (x, y) to ax + by = c has the form in part (b), given ax 0 + by 0 = c. Proof. Claim 1-(Proof by contradiction) Suppose gcd(a, b) c and there IS a solution (x, y). Then by the previous Lemma, since gcd(a, b) a and gcd(a, b) b, then gcd(a, b) ax + by = c (these two implication arrows going opposite directions means there is a contradiction in the logic, so it must 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 and t such that: as + bt = gcd(a, b) therefore; giving a particular solution; ask + btk = k gcd(a, b) = c. x 0 y 0 = sk = tk. Given this particular solution, the general solution now follows: ( ) ( ) b a x 0 + gcd(a, b) k a + b y 0 gcd(a, b) k = ax 0 + abk gcd(a, b) + by 0 = ax 0 + by 0 + abk gcd(a, b) = ax 0 + by 0 = c bak gcd(a, b) abk gcd(a, b) Claim 3-Suppose ax + by = ax 0 + by 0 = c then, ax + by ax 0 by 0 = c c = 0 = ax ax 0 + by by 0 = a(x x 0 ) + b(y y 0 ) a(x x 0 ) = b(y 0 y) But recall that d = gcd(a.b), by definition, divides a and b,and since it is the greatest common divisor, it makes sense that gcd( a, b ) = 1 (Euclid s lemma). d d

16 1.3 LINEAR DIOPHANTINE EQUATIONS IN Z 11 a d (x x 0) = b d (y 0 y) gcd( a d, b d ) = 1 a d (y 0 y) y 0 y = a k, for some k Z d a d (x x 0) = b d a d k (x x 0) = b d k x = x 0 + b d k y = y 0 a d k Example 1.7 Solve the Diophantine equation 42823x y = 51. The first step in solving a Diophantine equation, is to check to see if it is even solvable. To do that, it is necessary to use the previous listed theorems and find the gcd(a, b) and then see if the gcd divides the c terms. In Example 2, 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 integers x and y such that: 42823x y = 17 To do this, a series of substitutions are required. Going back to Example 2, recall the following steps in finding the gcd, but this time, in each case, solve for the remainder as follows: = (1) ( 6) = = (1) ( 1) = = (1) ( 2) = 289

17 12 CHAPTER 1 Z THE INTEGERS 2040 = (1) + 289( 7) = 17 Now, beginning with the last equation, a long string of back substitutions and distributive multiplication must be performed as follows: 2040(1) + 289( 7) = (1) + [4369(1) ( 2)]( 7) = 2040(1) ( 7) (14) = 2040(1 + 14) ( 7) = 2040(15) ( 7) = (15) ( 7) = [6409(1) ( 1)](15) ( 7) = 6409(15) ( 15) ( 7) = 6409(15) ( 22) = (15) ( 22) = 6409(15) + [42823(1) ( 6)]( 22) = 6409(15) ( 22) (132) = ( 22) (147) = 17 Hence, x = 22 and y = 147 for the equation 42823x+6409y = 17. Now to solve the original problem, since 51 = 3 17, we can multiply both sides of the gcd equation by 3 and distribute it in to the x and y solutions as follows: 3[42823( 22) (147)] = (3 ( 22)) (3 147) = ( 66) (441) = 51 The solution to the original problem is: x = 66 and y = 441.

18 1.4 WHAT IS THE BEST THE INTEGERS CAN BE?-RINGS 13 Definition 1.14 From now on, gcd(a, b) = (a, b). Example 1.8 Can solutions to the above problem be found in which both are positive? Obviously since both and 6409 are positive and greater than 51, it will not be possible, but another way to see this is to apply the infinite solution formulae: x = x 0 + b (a, b) k y = y 0 a (a, b) k x = k = k 17 y = k = k k > 0 37k > 66 k > k > > 2519k k < Since k must an integer, this wil be impossible, thus we conclude that both solutions cannot be positive. Lecture Problem 1.2 If solvable, find all negative integer solutions, 6x + 15y = What is the best the Integers can Be?- Rings The best structure (more on this word best later) that can be constructed directly from the integers is a ring.

19 14 CHAPTER 1 Z THE INTEGERS Definition 1.15 A Ring (R, +, ) is an Abelian group (R, +) with a secondary 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 all elements in R. Definition 1.16 A ring (R, +, ) is a COMMUTATUVE ring iff is a commutive operation. Definition 1.17 A ring (R, +, ) is a RING WITH UNITY iff has an identity element (often referred to as 1) Example 1.9 (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 did have inverses for all non-zero elements, and comprised our first example of the following structure: Definition 1.18 A Field (F, +, ) is comprised of two Abelian groups, (F, +) and (F \ {0}, ),related by the DISTRIBUTUVE PROPERTY, with the condition that 0 1, i.e., the identities are distinct. Example 1.10 Some commonly known fields: (Q, +, ) (R, +, ) (C, +, ). This book has several major themes: Appreciate the connections between the Fundamental Theorems of Arithmetic and Algebra; Learn the pedagogical power of rich-problem topics.

20 1.4 WHAT IS THE BEST THE INTEGERS CAN BE?-RINGS 15 Understand how fields can be constructed from the rings Z and Z[x]. Appreciate the importance of primary instruction and arithmetic to the secondary grades and algebra. In this respect, the next chapter revisits the rational numbers to go over several area models which also can be applied to introductory algebra.

21 16 CHAPTER 1 Z THE INTEGERS 1.5 Chapter 1 Exercises Exercise 1 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 2 Solve the Diophantine equation 56x + 72y = 40. Exercise 3 Solve the Diophantine equation 24x + 138y = 18. Exercise 4 Solve the Diophantine equation 85x + 30y = 215. Exercise 5 One of the Diophantine equations below is solvable. solution to the solvable equation: Find the 195x + 221y = 12 Exercise 6 Reduce the fraction x + 123y = 12 by finding (7739, 31937). Exercise 7 Reduce the fraction by finding (96496, ) Historical Applications Exercise AD Alcuin of York- A hundred bushels of grain are distributed among 100 people in such a way that each man receives 3 bushels, each woman 2 bushels and each child a half a bushel. Find a solution for how many men, women and children there are. How many possible answers are there?

22 1.5 CHAPTER 1 EXERCISES 17 Exercise 9 850AD-Mahavira-There were 63 equal piles of plantain fruit put together with seven single plantains. They were divided evenly among 23 travelers. How many plantains did each traveler get? [Hint: Consider 63x + 7 = 23y.] Figure 2 Exercise AD-Yen Kung-We have an unknown number of coins which were put on a number of strings in equal amounts (see Fig.2). If you make 77 strings of them you are 50 coins short, but if you make 78 strings then you have the exact number of coins. Find the number of coins [Hint: Consider 77x + 50 = 78y.] Exercise 11 Find all positive integer solutions to: 172x + 20y = Exercise 12 Find a solution to the following Diophantine equation, having an x-value between 17 and 20 : 56x + 72y = 16. Exercise 13 Using a prime factor calculator on the web, such as: http : // factors.php determine the number of positive factors for 4, 988, 552, 244.

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24 Chapter 2 Induction and Algebraic Thinking Algebraic reasoning consists of three basic components: 1. GENERALIZATION- Mathematical generalization involves a claim that some property or technique holds for a large set of mathematical objects or conditions. The scope of the claim is always larger than the set of individually verified cases. Example 2.1 A young student, after seeing enough individual cases like = 20 or = 114, could generalize the idea that the sum of ANY two even integers is an even integer. To prove this is the case in general requires the powerful generalizing notation found in algebra, such as characterizing even integers as integers of the form 2k, for some integer k. Theorem 2 The sum of any two even integers is even. Proof. Let 2k and 2l be two even integers. Then 2k + 2l = 2 (k + l) which is also an even integer. Note that the distributive property was used to factor out the 2. What generalization can you make about the following sums: = 12, = 156, = 22? Lecture Problem 2.1 Prove that the sum of any two odd integers is even. 19

25 20 CHAPTER 2 INDUCTION AND ALGEBRAIC THINKING 2. EQUALITY- The meaning of the = sign. Equality also relates to one of the most important aspects of algebraic thinking, substitution and a fancy term called encapsulation, which relates to the ability to treat complex representations as an object. Example 2.2 It is usually diffi cult for beginning algebra students to make use of structure and understand how the following equivalent expression can be gotten so easily. (2x + 1) (3x 2) + (2x + 1) = (2x + 1) (3x 1). 3. UNKNOWN QUANTITIES-The variable concept is what is usually associted with the term algebra. Using algebraic notation to work with unknown quantities is what makes algebra so powerful, but also so abstract and diffi cult to understand, when coming from learning arithmetic. It is the generalizing character of working with notation, along with the notion of equality, in which the three aspects of algebraic thinking come together, so that equations can be solved, and unknowns can be found. In the next section, a powerful proof technique will be studied, which is used to prove formulas generalized from seeing patterns. 2.1 Proof by Mathematical Induction Now that we have a better idea of what the integers are, we will investigate a technique for proving statements, P (n), about the strictly positive integers, the natural numbers: N = {1, 2, 3,...}. Mathematical induction to prove P (n) consists of steps: 1. The STATEMENT you want to prove holds true, P (n). 2. The BASE CASE step, show P (1) is true, unless the statement P (n) is defined for natural numbers such as n 2 (then you would start with proving P (2) because P (n) is probably undefined at n = 1).

26 2.1 PROOF BY MATHEMATICAL INDUCTION The ASSUMPTION step, assume P (k) is true. 4. The INDUCTION step, show that P (k + 1) is true assuming P (k) is true. Example 2.3 About the easiest proof one might start off with in induction is a statement such as, P (n) : n < n + 1. proof: 1. P (1) : Indeed it is true that 1 < = 2 2. Assuming P (k) is true then k < k + 1. Adding 1 to each side, an equation and inequality preserving property, gives k + 1 < k (k + 1) < (k + 1) + 1 P (k + 1)

27 22 CHAPTER 2 INDUCTION AND ALGEBRAIC THINKING Example 2.4 Another usual first example is Gauss Formula, proof: P (n) : n = n(n + 1) P (1) : 1? = 1(1+1) 2... yes, 1 = = 1 2. Assuming P (k) is true then k = k(k+1) (This is known 2 as the inductive hypothesis). We need to prove that P (k + 1) is true. A useful thing to do is to write the statement P (k + 1) while including the statement P (k) in it on the LHS, (k + 1)(k + 2) P (k + 1) : k + k + 1 =. 2 Now, mold the LHS into the RHS by substituting the RHS of the inductive hypothesis, P (k), into the LHS of P (k + 1). k(k + 1) k + k + 1 = 2 = k2 + k k + 1 = k2 + k + (k + 1) (k + 1) = k2 + 3k + 2 = 2 2 (k + 1)(k + 2) 2 Lecture Problem 2.2 Prove by induction P (n) : (3n 2) = n(3n 1). 1 2 The previous example and lecture problems demonstrate a particular style of induction proof that molds the LHS of P (k + 1) into its RHS. The next style of induction proof, typically seen in divisibility statements, directly works on molding the inductive hypothesis P (k) into P (k + 1).

28 2.1 PROOF BY MATHEMATICAL INDUCTION 23 Example 2.5 Prove by induction P (n) : 9 n 1 is divisible by 8, for all n N. proof: 1. P (1) : = 8 and 8 divides 8 2. Assuming P (k) is true then 8 9 k 1, then by definition, 9 k 1 = 8m for some natural number m N. To obtain P (k + 1) one needs, P (k + 1) : 9 k+1 1 = 8r, for some natural number m N. A way to at least make a 9 k+1 appear from the inductive hypothesis is to multiply both sides by 9, 9 (9 k 1 ) = 9 8m 9 k+1 9 = 9 k = 9 8m 9 k+1 1 = 9 8m + 8 = 8 (9m + 1) 8 9 k+1 1 Definition 2.1 ALGEBRAIC PATTERNS are sequences of numbers with patterns based on addition or subtraction. Lecture Problem 2.3 Determine the last two numbers in this algebraic pattern: 47, 43, 40, 38, 37, 33,,. Definition 2.2 GEOMETRIC PATTERNS are sequences of numbers with patterns based on multiplication or division. Lecture Problem 2.4 Determine the last two numbers in this geometric pattern: 15, 3, 3,,. 5

29 24 CHAPTER 2 INDUCTION AND ALGEBRAIC THINKING 2.2 Chapter 2 Exercises Exercise 14 Prove by induction P (n) : (6n 5) = n(3n 2). Exercise 15 Prove by induction P (n) : (2n 1) = n 2. Exercise 16 Prove by induction P (n) : n 1 = 1 4 (5n 1). Exercise 17 Prove by induction P (n) : n(n + 1) = n(n + 1)(n + 2). 1 3 Exercise 18 Prove by induction P (n) : 1 n. 2n (2n 1) (2n+1) = Exercise 19 Prove by induction P (n) : 3 2n 1 is divisible by 8, for n 0. Exercise 20 Prove by induction P (n) : 3 2n 1 +1 is divisible by 4, for n N. Exercise 21 Prove by induction P (n) : 2 n n+1 is divisible by 7 for n 0. Exercise 22 Prove by induction P (n) : 6 n(n + 1)(n + 2) for n N. Exercise 23 Prove by induction P (n) : 8 n 3 n is divisible by 5 for n N. Exercise 24 Given the sequence 12,, 72, 216, 432, 1296, what number goes into the box? Is this pattern algebraic or geometric?

30 2.2 CHAPTER 2 EXERCISES 25 Classroom Management Corner Exercise 25 Draw a 1-panel comic of a humourous classroom management situation. The above is an example. Exercise 26 After viewing the Youtube video Unconventional Teacher Tactics that Quell Bad Behavior, write a 1-page minimum word-processed reflection for what you liked or dis-liked about the suggestions, and a few examples for how it could relate to your future teaching and/or past experiences observing other classrooms.

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32 Chapter 3 Exponents and the Binomial Theorem This chapter discusses the connections between the different sets of numbers in mathematics and their common algebraic representations. 3.1 Set Theory Basics Numbers may seem like the basic building blocks of mathematics, but really the basic building blocks are SETS. A set is a collection of objects called the elements or members of the set. These objects could be anything conceivable including numbers, letters, colors, even sets themselves! However, none of the objects of the set can be the set itself. We discard this possibility to avoid running into Russell s Paradox, a famous problem in mathematical logic unearthed by the great British logician Bertrand Russell in Definition 3.1 A SET is a collection of objects. Definition 3.2 For sets A and B in some universal set U, A B = {x U x A or x B}. This is called the union of A and B. 27

33 28CHAPTER 3 EXPONENTS AND THE BINOMIAL THEOREM Definition 3.3 For sets A and B in some universal set U, A B = {x U x A and x B}. This is called the intersection of A and B. Definition 3.4 A = {x U x / A}. the set A. This is called the complement of Definition 3.5 For sets A and B in some universal set U, A\B = {x U x A and x / B}. This is called A complement B. Definition 3.6 (Subset Concept) A B means that x A x B. Definition 3.7 (The Empty Set) = { }. Remark 3.1 { } = {{}}.

34 3.2 NUMBER SETS Number Sets Here are some important number sets used in mathematics: N = {1, 2, 3,...} is the set of counting numbers, or natural numbers. W = {0} {1, 2, 3,...} = {0, 1, 2, 3,...} is the set of whole numbers. Z = {... 3, 2, 1, 0, 1, 2, 3,...} is the set of integers. Q = { p p, q Z and q 0} is the set of rational numbers. This set q can also be described as the set of all terminating or non-terminating REPEATING decimal numbers. I is the set of all irrational numbers. This set also be described as the set of all non-terminating NON-REPEATING decimal numbers.

35 30CHAPTER 3 EXPONENTS AND THE BINOMIAL THEOREM R is the set of real numbers. In set notation: R = Q I. C is the set of complex numbers. In set notation: C = {a+bi a, b R and i = 1}. Definition 3.8 A prime number, p, is a natural number with exactly two divisors, 1 and p. Definition 3.9 A composite number is a natural number that is not prime. 3.3 Justifications of Exponent Properties The following theorem can be proven given the following definitions Theorem 3 Given a m = a a a a m times a 0, then a = 1, then: a for any natural number m and if a m a n = a m+n Proof. a m a n = a a a a n times ( ) ( ) a a a a a a a a m times n times = a a a a = a m+n m+n times = a a a a m times Using the above proof as a model, try and do the next two on your own! (a b) m = a m b m

36 3.3 JUSTIFICATIONS OF EXPONENT PROPERTIES 31 (a m ) n = a mn a = a 1/2 Proof. a is defined as a number such that when it is squared, it results in ( ) 2 a. By the previous exponent property, since a = a 2 2 = a 2 2 = a 1 = a, then it follows that a = a 1/2 n a = a 1/n am a n a 0 = 1 = a m n Proof. This follows immediately from the previous property since, assuming a 0, then am a m = 1 = a m m = a 0 a m = 1 a m Proof. a m a m = a m+( m) = a 0 = 1 = am = a m 1,so: a m a m.a m a m = a m 1 a m a m = 1 a m Remark 3.2 As a consequence of the properties (a b) m = a m b m and n a = a 1/n, one can simplify radical expressions. Example = 16 3 = 16 3 = 4 3. Example = = =

37 32CHAPTER 3 EXPONENTS AND THE BINOMIAL THEOREM 3.4 p is an Irrational Number Theorem 4 p is irrational for any prime number Proof. (by contradiction) Suppose that p = a for some positive integers b a and b with b 0. Then after squaring both sides, p = a2 b p 2 b2 = a 2. But any prime number in the factorization of any natural number such as m must occur an EVEN number of times in the factorization of m 2 (Why?) Therefore, since a 2 = p b 2, then it must be the case that the prime number p is in the factorization of a 2 an ODD number of times, which is a contradiction, so it must be impossible to write p = a, thus it is concluded b that p is irrational 3.5 The Binomial Theorem Issac Newton ( ) Some Brief History The Binomial theorem is one of the most important algebraic formulas for the history of Calculus. In some respect, we have the plague to thank for the binomial theorem! In 1665, plague was raging in England, and Isaac Newton, a new (and undistinguished) graduate of the University of Cambridge, was forced to spend most of the next two years in the relative safety of his family s country manor in Woolsthorpe.

38 3.5 THE BINOMIAL THEOREM 33 It turned out that solitude and free time was just the stimulous Newton s creative brain needed. In that 18-month period of retreat, he came up with his proof and extension of the binomial theorem, invented calculus (which he called his "method of fluxions"), discovered the law of universal gravitation, and proved that white light is composed of all colors. All of this before the age of 25! Newton was not the first to describe a formula for binomial expansions, or multiplying out any expression of the form (a + b) n. We know, for example, that an Islamic mathematician named al-karaji (d. 1029) constructed a table of binomial coeffi cients up to (a + b) 5 (that is, Pascal s triangle as seen above), and later Muslim mathematicians credited him with discovering the formula for the expansion of (a + b) n. Furthermore, in a now lost work, Omar Khayyam ( ) apparently gave a method for finding nth roots based on the binomial expansion and binomial coeffi cients. Ancient Indian and Chinese mathematicians also knew the binomial theorem. And in Europe, already a century before Newton s birth, Blaise Pascal s Treatise on the Arithmetical Triangle provided a handy way to generate binomial coeffi cients. All of these methods for binomial expansion, however, work only for positive integer values of n. The Binomial Theorem states that for real or complex a, b, and nonnegative integer n, ( n (a + b) n = a n + 1 n ( n = i where, i=0 ) a n 1 b + ) a n i b i. ( ) ( ) n n a n 2 b ab n 1 + b n 2 n 1 ( n ) k = n! k!(n k)! is a binomial coeffi cient. In other words, the coeffi cients when (a + b) n is expanded and like terms are collected are the same as the entries in the nth row of Pascal s Triangle above. Example 3.3 (a + b) 5 = a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5, with coeffi cients 1 = ( 5 0),5 = ( 5 1),10 = ( 5 2) etc..

39 34CHAPTER 3 EXPONENTS AND THE BINOMIAL THEOREM Proofs There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof, which will be the focus of this text; but, since we have discussed mathematical induction in the previous chapter, the inductive proof will be shown, although it relies heavily upon knowledge of the following property of the n formula which k is intimatey connected to generating Pascal s Triangle: The Binomial Theorem proof by Mathematical Induction: Proof. The proof is by induction on n. When n = 0, we have (a + b) 0 = 1 and 0 i=0 ( ) n a n i b i = i = 1. ( ) 0 a 0 0 b 0 0 Therefore, the statement is true when n = 0. Suppose now that the statement is true for some integer n = k 1 where k > 1. Then, (a + b) n = (a + b)(a + b) n 1 n 1 ( n 1 = (a + b) i i=0 n 1 ( n 1 = i i=0 n 1 ( n 1 = i i=0 ) a n i b i + ) a n i b i + ) a n 1 i b i n 1 ( n 1 i i=0 n ( n 1 i=1 i ) a n i b i+1 ) a n 1 i b i

40 3.5 THE BINOMIAL THEOREM 35 n 1 ( n 1 = a n + i i=1 n 1 [( ) n 1 = a n + + i i=1 n 1 ( ) n = a n + a n i b i + b n i = n i=0 i=1 ( n i ) a n i b i. ) a n i b i + n 1 ( n 1 i 1 i=1 ( n 1 i 1 )] a n i b i + b n ) a n 1 i b i + b n Therefore, thestatement is true when n = k + 1,and so it is true for all n 0 Lecture Problem 3.1 Use Pascal s Triangle to expand: (2a + b) 6. Lecture Problem 3.2 Find the seventh term in the expansion of :. ( x 2 y ) 15. Lecture Problem 3.3 Find the term with no m, in the expansion of: ( m 1 m 3 ) 12. Lecture Problem 3.4 Find the middle term in the expansion of: ( m 1 m 3 ) 12. Exercise 27 Find a polynomial that has α = as a root.

41 36CHAPTER 3 EXPONENTS AND THE BINOMIAL THEOREM 3.6 Chapter 3 Exercises Exercise 28 Describe each of the following by drawing seperate shaded diagrams on a diagram such as above. A A B A B (A B) (A B) A B A B U\[(A\B) (B\A)] Exercise 29 How many positive factors are in the number ? many integer factors does the number have? How Exercise 30 Simplify: ( 2x 6 ) (3x 5 ) 2. Exercise 31 Simplify: (A, B 0) 6A 3 B 7 3A 4 B 6.

42 3.6 CHAPTER 3 EXERCISES 37 Exercise 32 Simplify: 216 2/3. Exercise 33 Simplify: 8 2/3. Exercise 34 Types of Numbers Worksheet. Exercise 35 Use Pascal s Triangle to expand: (2a + b) 6. Exercise 36 Find the seventh term in the expansion of :. ( x 2 y ) 15. Exercise 37 Find the term with no m, in the expansion of: ( m 1 m 3 ) 12. Exercise 38 Find the middle term in the expansion of: ( m 1 m 3 ) 12. Exercise 39 Find a polynomial that has α = as a root. Exercise 40 Find a polynomial that has α = and β = as roots. Exercise 41 Prove 3 is an irrational number.

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44 Chapter 4 Exploring Quadratic Equations 4.1 Parabolas and Interpolation Just as two points determine a unique line, three points determine a unique parabola, or quadratic equation. In this section, we will find the equations of parabolas given three points by using matrices to solve a system of equations. But first a few definitions and principles about matrices: Definition 4.1 The n n identity matrix is a matrix with n rows and n columns with 1 s on the main diagonal and zeros everywhere else The Three Principles of Row Reduction for Solving a System of Equations 1. Rows can be added to other Rows without changing the solution to the system. 39

45 40 CHAPTER 4 EXPLORING QUADRATIC EQUATIONS 2. A Row can be multiplied by a non-zero number (scalar) without changing the solution to the system. 3. Rows can be switched without changing the solution to the system. The goal in reducing a system of equations for an augmented matrix is to obtain the goal matrix for the coeffi cient matrix by using combinations of the 3 principles of row reduction a b y z Example 4.1 Given the points (1, 7), ( 1, 9), and (2, 15) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. Solution: First, three equations in the unknown variables a, b, and c are formed upon substitution of the (x, y) given ordered pairs. Then an augmented matrix is used and the three principles of row reduction are employed to obtain the identity coeffi cient matrix as described above. (1, 7) a (1) 2 + b (1) + c = 7 ( 1, 9) a ( 1) 2 + b ( 1) + c = 9 (2, 15) a (2) 2 + b (2) + c = 15 a + b + c = 7 a b + c = 9 4a + 2b + c = 15

46 4.2 TWO FORMS OF A QUADRATIC FUNCTION 41 Begin Row Reduction on the 3 4 system matrix by focusing on creating zeros below the pivot 1 s, after making sure one has a 1 in the first pivot position in the upper lefthand corner R1+R2 1 3 R R R1+R3 Now for back substituion zero out above the pivots R2+R1 R3+R1 a = 3, b = 1 and c = R2+R3 Final answer: y = 3x 2 x + 5. Lecture Problem 4.1 Given the points (1, 2), ( 1, 14), and (3, 10) find the equation, in the form y = ax 2 +bx+c, of the parabola through those points. Lecture Problem 4.2 Given the points (1, 2), ( 2, 5), and (2, 9) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. Lecture Problem 4.3 Given the points (2, 8), ( 2, 12), and (3, 12) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. 4.2 Two Forms of a Quadratic Function A quadratic function has a graph that is a parabola.

47 42 CHAPTER 4 EXPLORING QUADRATIC EQUATIONS The General Form We have already introduced this form in the previous section. y = ax 2 + bx + c The Vertex or Standard Form This form is best for easily sketching the graph of a parabola with vertex at (h, k). y = a(x h) 2 + k As explored in lecture, we should discover that (x h) represents the HORIZONTAL SHIFT of h units right, if h > 0 and k represents a vertical shift of the graph in the direction of k. The amplitude of the parabola is a. a > 0 is CONCAVE UP. a < 0 is CONCAVE DOWN. Note: The magnitude of a determines how thin the parabola is: a > 1 means the parabola is narrower than the standard base-graph parabola y = x 2. 0 < a < 1 means the parabola is wider than the standard base-graph parabola y = x 2. The method for finding the vertex form given the standard or general form is by completing-the-square. Lecture Problem 4.4 Find the vertex form for the quadratic function y = x 2 4x + 7l and sketch its graph. Lecture Problem 4.5 Find the vertex form for the quadratic function y = 2x 2 + 4x + 5 and sketch its graph. Lecture Problem 4.6 Find the vertex form for the quadratic function y = 1 2 x2 x 6 and sketch its graph.

48 4.3 FACTORING TRINOMIALS Factoring Trinomials The ac-cloud method for factoring ax 2 + bx + c over the Integers Lecture Problem 4.7 Factor the given trinomial, 22x 2 31x 3. Lecture Problem 4.8 Factor the given trinomial, 6x x 28. Lecture Problem 4.9 Factor the given trinomial, 48x x Projectile Motion The parabola is one of the most important mathematical shapes for physics. Just throw a ball in the air at a bit of an angle to see why Equations of Motion Vertical Equations y = 1 2 a yt 2 + v 0y t + y 0 Horizontal Equations v y = v 0y + a y t x = 1 2 a xt 2 + v 0x t + x 0 v x = v 0x + a x t

49 44 CHAPTER 4 EXPLORING QUADRATIC EQUATIONS Lecture Problem 4.10 A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. Lecture Problem 4.11 A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table s edge and the ball s landing location. Lecture Problem 4.12 A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the longjumper.

50 4.5 CHAPTER 4 EXERCISES Chapter 4 Exercises Exercise 42 Given the points (1, 6), (2, 23), and (3, 52) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. Exercise 43 Given the points (2, 3), ( 2, 5), and (3, 10) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. Exercise 44 Given the points (1, 8), ( 2, 17), and (2, 17) find the equation, in the form y = ax 2 + bx + c, of the parabola through those points. Exercise 45 Find the vertex form for the quadratic function, y = 2x 2 + 8x + 13 and sketch its graph. Exercise 46 Find the vertex form for the quadratic function, y = 1 2 x2 3x 5 2 and sketch its graph. Exercise 47 Factor the given trinomial, Exercise 48 Solve: 20x x 15. 8x x 21 = 0. Exercise 49 An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object s height s at time t seconds after launch is s(t) = 4.9t t , where s is in meters. When does the object strike the ground? Exercise 50 A daredevil tries to jump a canyon of width 10 m. To do so, he drives his motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed is necessary to clear the canyon? Exercise 51 A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.