Deepening Mathematics Instruction for Secondary Teachers: Algebraic Structures

Transcription

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

2 Contents Preface ix 1 Z The Integers What are the Integers? The Euclidean Algorithm History Division Algorithm Proof of the Euclidean Algorithm Linear Diophantine Equations in Z Solving ax + by = c Linear Diophantine Equation Proofs Chapter 1 Exercises v

3 Preface This text is for students who want to become secondary mathematics teachers. The theme is to show the structural algebraic connections between sets of numbers and sets of polynomials. One approach to this is the study of Diophantine equations in multiple settings, such as integers, integers modulon and single variable polynomials. This approach also emphasizes the role of questioning and justification in learning mathematics, as well as rich-problem lesson design. 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. Lance Burger January, 2014 ix

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5 Chapter 1 Z The Integers 1.1 What are the Integers? 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 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

6 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 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 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 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

7 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 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 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.

8 4 CHAPTER 1 Z THE INTEGERS Proof of the Euclidean Algorithm Proof. Given integers b and c with c > 0, we make repeated applications of the Division Algorithm to obtain a series of equations: b = cq 1 + r 1, where 0 < r 1 < c. This continues with, c = 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(b, c) = r j, the last nonzero remainder. Variables x and y in gcd(b, c) = bx + cy is obtained by writing each r j as a linear combination b and c. The proof is very simple. The chain of equations is obtained by dividing c into b, r 1 into c, 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 < c, however, for the Euclidean Algorithm, it is simply 0 < r < c. The reason being is that if r = 0, then the series of equations would stop at b = cq 1, implying the gcd(b, c) = c. Last, we need to prove that r j is the greatest common divisor or b and c. First, we observe the following: gcd(b, c) = gcd(b cq 1, c) = gcd(r 1, c) = gcd(r 1, c 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(b, c) = gcd(r j 1, r j ) = gcd(r j, 0) = r j. Concluding the proof that r j = gcd(b, c).

9 1.2 THE EUCLIDEAN ALGORITHM 5 Corollary 1 As a consequence of the above proof, back substitution can be performed, similar to the following example, so that a linear combination can always be found such that: bx + cy = gcd(b, c) Example 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 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 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.

10 6 CHAPTER 1 Z THE INTEGERS Reflection Question(s) 1 Based on the previous definition does 0 12 make any sense? Definition 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 8 A PRIME number is a positive integer p with exactly two factors, p and 1. Example 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 How many positive factors does 5400 have? Definition 9 Integers a and b are RELATIVELY PRIME iff gcd(a, b) = 1.

11 1.2 THE EUCLIDEAN ALGORITHM 7 Definition 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 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 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

12 8 CHAPTER 1 Z THE INTEGERS 1.3 Linear Diophantine Equations in Z Solving ax + by = c Definition 11 A DIOPHANTINE EQUATION is any equation in which the solutions are over the integers. Remark 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 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?

13 1.3 LINEAR DIOPHANTINE EQUATIONS IN Z Linear Diophantine Equation Proofs Preliminaries: Definition 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 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 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 2 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.

14 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 + 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 y 0 ) But recall that gcd(a.b), by definition, divides a and b.

15 1.3 LINEAR DIOPHANTINE EQUATIONS IN Z 11 Example 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) = = (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) = 17

16 12 CHAPTER 1 Z THE INTEGERS 6409(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. Lecture Problem 2 If solvable, find integer solutions for 6x + 12y = 30. Lecture Problem 3 If solvable, find integer solutions for 18x+72y = 55. Lecture Problem 4 If solvable, find integer solutions for 60x+770y = 500.

17 1.4 CHAPTER 1 EXERCISES 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 gcd(7739, 31937). Exercise 7 Reduce the fraction by finding gcd(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. How many men, women and children are there?

18 14 CHAPTER 1 Z THE INTEGERS 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.]