Fundamental Mathematical Concepts Math 107A. Professor T. D. Hamilton

Fundamental Mathematical Concepts Math 107A Professor T. D. Hamilton January 17, 2007

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Contents 1 Set Theory 7 What is a set?.......................................... 7 Describing a Set......................................... 8 2 vs............................................... 9 Set Operations.......................................... 13 Union............................................ 13 Intersection......................................... 13 Set Di erence........................................ 14 Complement......................................... 14 Relationships........................................ 15 Properties.......................................... 15 Venn Diagrams....................................... 17 Equivalent Sets.......................................... 22 Finite vs. In nite Sets................................... 24 Chapter 1 Exercises....................................... 24 2 Arithmetic of the Natural Numbers 27 Development of the Natural Numbers ***........................... 27 Developing a New Number System (the fen system) ***................... 28 Activity Fenland Economics I................................. 28 Addition.............................................. 29 De nition.......................................... 29 Algorithm ***........................................ 30 Activity Cuisinaire Rods.................................... 31 Subtraction............................................ 31 A First Model Take-Away................................ 31 De nition.......................................... 32 Algorithm ***........................................ 32 Models for Subtraction................................... 32 Activity Place Value Game.................................. 34 Activity Base 10 Blocks.................................... 35 Converting between Fen and Ten................................ 36 Multiplication........................................... 38 De nition.......................................... 38 Array Model of Multiplication ***............................ 38 Algorithm ***........................................ 38 Lattice Algorithm ***................................... 38 3

4 CONTENTS Activity Fenland Economics II................................ 38 Division (without remainder).................................. 39 De nition.......................................... 39 Models for Division..................................... 41 Division with Remainder..................................... 42 De nition.......................................... 42 Algorithm ***........................................ 43 Mental Arithmetic and Estimation ***............................. 43 Chapter 2 Exercises....................................... 43 3 Number Theory 51 Divisibility............................................ 51 Primes............................................... 52 Fundamental Theorem of Arithmetic.............................. 56 Number of Divisors ***..................................... 57 Tests for Divisibility....................................... 57 Divisibility by 2, 5 and 10 ***............................... 57 Divisibility by 3 and 9 ***................................. 57 Divisibility by 11 ***.................................... 57 Divisibility by 2 r ***.................................... 57 Divisibility Rules for Composites ***........................... 57 Greatest Common Divisor (GCD) and Least Common Multiple (LCM).......... 57 De nitions.......................................... 57 Set Intersection Method.................................. 58 Prime Power Representation Method ***........................ 58 Euclidean Algorithm Method ***............................. 58 Activity Interesting Numbers................................. 59 Chapter 3 Exercises....................................... 60 4 The Integers 65 Opposites............................................. 65 Representations of the Integers................................. 66 Activity Colored Counters................................... 67 Subtraction as Addition..................................... 69 Addition of Integers....................................... 70 Absolute Value.......................................... 72 De nition.......................................... 72 Geometric Interpretation.................................. 72 Inequalities............................................ 74 De nitions.......................................... 74 Geometric Interpretation.................................. 75 Multiplication of Integers.................................... 76 Division of Integers........................................ 77 Chapter 4 Exercises....................................... 78

Preface This course is designed to help to prepare you to teach mathematics in the elementary school classroom. For this reason, we will be less concerned with how to do mathematics and more concerned with why we do it the way we do. That is, why are the methods that we learned (and teach) valid. For example, I assume that you know how to compute 356 57: In this course we will discuss why the technique you used to compute 356 57 is valid, that is, why you get the right answer. We ll also discuss other techniques and why they work. In the end you should be able to teach your students how to multiply numbers and you should be able to present the material in such a way that your students will understand why they are doing what they are doing. That way your students will come away with a better understanding of mathematics. They will be able to explain what they are doing without having to say, I do it that way because my teacher said so. 5

6 PREFACE

Chapter 1 Set Theory Set theory is essentially just the study of collections of objects. It shows up throughout the grade school mathematics curriculum. Consider how you were rst taught to add 3 + 2: You probably did something like this: if I have 3 apples and someone gives me 2 more apples, how many apples will I have? In this example, you are taking a set with 3 objects and a set with 2 objects and nding the size of the union (see de nition on page 13) of these two sets. We will come back to this idea on page 29 when we discuss addition in more detail. What is a set? We will not have a rigorous de nition of the word set. Instead, we will use the word set to describe any collection of objects. The objects in the set are usually referred to as elements of the set. A set does not have to be mathematical. One important set is the set with no elements. This set is called the empty set and is often denoted by the symbol ;: It is simply an empty collection. It is important to remember that the empty set is not nothing; it is something, it is a set with no elements, i.e. a set with nothing in it. Consider, for example, the set of all students at California State University Sacramento who commute daily from New York City. This is a set, but there is certainly nobody in the set. Thus, this set is an empty set. Another example would be to take the set of all numbers x such that x 6= x: This set is also an empty set. De nition 1 We say that two sets are equal if they contain exactly the same elements. The two empty sets mentioned in the previous paragraph are equal sets; they each contain no elements, so, of course, they contain the same elements. Thus, the set of all students at California State University Sacramento who commute daily from New York City is equal to the set of numbers x such that x 6= x: In fact any empty set is equal to any other empty set. This is why we are able to refer to the empty set rather than an empty set. 7

8 Set Theory Now let s consider an example of equal sets that aren t empty. Example 2 Let A be the set of all numbers x such that x 2 = 1: Let B be the set of all numbers x such that jxj = 1: Notice that A contains the elements 1 and 1 and B contains the elements 1 and 1: Therefore, A = B: Describing a Set There are many ways of describing a set. So far we have been describing sets in words. That is, we have simply described the conditions necessary for an object to be an element of the set. Another option is to describe a set in roster notation. A roster is simply a list and, in roster notation, you simply list the elements of the set inside set brackets, f g : Example 3 Recall the set A from the previous example. In words we described A as the set of all numbers x such that x 2 = 1: In that example we said that A contains the elements 1 and 1: So, if we want to describe A in roster notation, we would write A = f1; 1g : Note that we could also put A in roster notation by writing A = f 1; 1g : Notice that if the set is a (small) nite set and you know the elements of the set, then roster notation is a very simple way of describing the set. Since the empty set is a set with no elements, we can write the empty set in roster notation as f g : That is, ; = f g : In fact, this is the de nition of the empty set. De nition 4 The set f g ; the set without any elements, is called the empty set and can be denoted by the symbol ;: ; = f g Note that ; is not equal to f;g since f;g is a set (given in roster notation) with one element, ;: One analogy that helps in understanding the di erence between these two sets is to think of a set as a bucket with the objects of the set inside the bucket. So, f1; 1g would be a bucket with two objects in it, the numbers 1 and 1: Using this analogy, ; would be an empty bucket while f;g would be a bucket containing an empty bucket (one bucket inside of another). When you describe a set in roster notation, it is easy to determine how many elements a set has and what those elements are. However, to make these decisions easy, you must describe the set without redundancy. For example, the set f1; 1; 1g is also equal to A: That is, A = f1; 1; 1g : Why is this true? Remember that, by de nition, two sets are equal if they contain exactly the same elements. In this case, A contains two elements, the numbers 1 and 1 and the set f1; 1; 1g also contains two elements, the numbers 1 and 1: So, A = f1; 1; 1g : If we write A as f1; 1; 1g ; we say that we have redundancies. When we write A = f1; 1g ; we say that we have put A in roster notation without redundancy. Example 5 Let B = 3; 6; 15 5 ; p 36; 18 15; 3 : Since 15 5 = 3 and p 36 = 6 and 18 15 = 3; we have B = f3; 6; 3; 6; 3; 3g : If we want to write B is roster notation without redundancy, we write B = f3; 6g : So, even though it originally appeared as if B had 6 elements, we found that B actually only contains 2 elements.

Set Theory 9 Example 6 Let C = f;; f g ; finishg A nal option is to describe a set in set-builder notation. The basic format for set-builder notation is as follows, fx : conditiong and is read the set of all x such that condition holds. The condition will change with the set. Think of it as the condition which determines residency in the set, that is, it is the condition that an object must satisfy to be in the set. Example 7 Let s write the set A, from the previous examples, in set-builder notation. Since A is the set of all numbers x such that x 2 = 1; the condition which determines residency in A is x 2 = 1 So, A = x : x 2 = 1 : To read this you would say A equals the set of all x such that x 2 = 1: Set-builder notation for a set is not unique. Notice that the set in the example above could also be written as fx : jxj = 1g (read the set of all x such that the absolute value of x is equal to 1 ) or fx : x = 1 or x = 1g (read the set of all x such that x is equal to 1 or x is equal to 1: ) 2 vs. We are now ready to introduce some notation. We will start with the symbol, 2 : This symbol is used to denote is an element of. Also, we use the symbol 62 to denote is not an element of. Example 8 Let A = f1; 1g : Then A has two elements, 1 and 1: So, we have 1 2 A and 1 2 A: Also, since 2 is not an element of A; we can write 2 62 A: The other notation,, is used to denote is a subset of. For this we need a de nition. De nition 9 Let X and Y be sets. We say that X is a subset of Y; written X Y; if every element of X is also an element of Y: If X is not a subset of Y; we write X 6 Y: Example 10 Let A = f1; 1g and let B = f 2; 1; 0; 1; 2g : Then A has two elements, 1 and 1: B has ve elements, 2; 1; 0; 1 and 2: So, every element of A is an element of B: So, A B: Notice however that not every element of B is an element of A since 2 2 B; but 2 62 A: Thus, B 6 A: Now, for a more complex example. In the next example, we will have a set whose elements are sets. This situation can seem confusing as rst, but it is really no di erent from the previous examples. Example 11 Let C = f;; 1; f1; 2g ; f2g ; 3; 4g : Before doing anything else, let s determine the elements of C: Since C is given in roster notation, we know that the elements of C are inside the outer set brackets and seperated by commas. I like to list the elements in a vertical list so that I can remove the seperating commas and see more clearly what the elements of C are.

10 Set Theory The elements of C are ; 1 f1; 2g There is no redundancy in this list, so C contains 6 elements. In words: the empty set the number 1 f2g 3 4 the set containing the numbers 1 and 2 the set containing the number 2 the number 3 the number 4. True or False: 1 2 C This is true since the number 1 is on the list of elements of C: True or False: 1 C This is false since the number 1 is not a set, so it cannot be a subset of C: True or False: 2 2 C This is false since the number 2 is not on the list of element of C: True or False: f2g 2 C This is true since f2g ; the set containing the number 2, is on the list of elements of C: True or False: f2g C This is false since the set f2g contains the element 2; but C does not contain the element 2: That is 2 2 f2g ; but 2 62 C: In order to have f2g C; we need to have every element of f2g be an element of C: True or False: f1; 2g C This is false since 2 2 f1; 2g but 2 62 C: In order for this to be true we would need every element of f1; 2g (that is: the number 1 and the number 2) to be an element of C: While the number 1 is an element of C; the number 2 is not. So we do not have every element of f1; 2g in C: So, f1; 2g 6 C: The concept of subset shows up in the way that we learn numbers as a child. We start with the simplest numbers, the counting numbers. The name for these numbers is the natural numbers. The symbol for the set of natural numbers is N: So, in roster notation, N = f1; 2; 3; 4; : : :g : Then we build up, by including 0, to the whole numbers. The symbol for the set of whole numbers is W: So, in roster notation, W = f0; 1; 2; 3; 4; : : :g : Notice that every element of the natural numbers, N; is an element of the whole numbers, W: So, N W:

Set Theory 11 Next, we learn about the negative numbers and build up to the integers. The symbol for the set of integers is Z: So, in roster notation, Z = f: : : ; 2; 1; 0; 1; 2; : : :g : Notice that every element of W is an element of Z; so we have W Z: (we also have N Z) Next we learn about fractions, the rational numbers. The symbol for the set of rational numbers is Q: It would be very di cult to give Q in roster notation, so we usually describe this set in set-builder notation, n a o Q = b : a 2 Z and b 2 Z and b 6= 0 : So, Q contains all numbers that can be written as a b where a and b are integers and b is not equal to 0. Notice that if n is an element of Z; then n = n 1, so n is an element of Q: Therefore, every element of Z is an element of Q: So, Z Q: We next move up to the real numbers. The symbol for the set of real numbers is R: The set of real numbers includes all of the numbers that we ve mentioned so far. It also contains numbers like p 2 and : The set of real numbers contains all numbers on the number line. Since all rational numbers are real numbers, we have Q R: Finally, we learn about the imaginary numbers. For these we usually start with the number i which is de ned by i 2 = 1, that is i = p 1: With this we can build the set of complex numbers, denoted C: The set of complex numbers includes all numbers of the form a + bi where a and b are real numbers. In set-builder notation this becomes C = fa + bi : a 2 R and b 2 Rg : Suppose r is a real number, then r = r + 0i which is an element of C: Thus, every element of R is an element of C: So, R C: Putting all of this together, we can describe the progressing of numbers learned in grade school in one line: N W Z Q R C: Arithmetic Sequences One type of set that we will have a special interest in is a set whose elements form an arithmetic sequence. De nition 12 A sequence of numbers is called an arithmetic sequence if each term of the sequence is found by adding a xed value (called the common di erence) to the previous term. Example 13 Each of the following sequences is an arithmetic sequence. 3,7,11,15,19,... (common di erence = 4) 8,13,18,23,... (common di erence = 5) 7,5,3,1,-1,-3,... (common di erence = 2) Example 14 Each of the following sequences is not an arithmetic sequence. 3,7,12,17,22,... (add 4 to term 1 to get to term 2, but add 5 to get from term 2 to term 3)

12 Set Theory 2,4,8,16,... (add 2 to term 1 to get to term 2, but add 4 to get from term 2 to term 3) We want to be able to take a set, given in roster notation, whose elements form an arithmetic sequence and give the set-builder notation for that set. Example 15 Let X = f3; 7; 11; 15; 19; : : :g : So, X is a set, given in roster notation, whose elements form an arithmetic sequence. If we want to nd a useful set-builder notation form of this set, we will need to nd a formula relating the elements of X: Notice that the rst element of this set is 3: After that, we add 4 to 3 to get 7: Then, we add 4 to 7 to get 11. Adding 4 to 7 is the same as adding 4 twice to 3. Notice that if we add 4 three times to 3 we get 15, the fourth element of the set. So, each element of this set is found by starting at the number 3 and adding some number of 4 s. That is, they all are of the form 3 + 4n: The rst element, 3; is 3 + 4 0: (n = 0) The second element, 7; is 3 + 4 1: (n = 1) The third element, 11; is 3 + 4 2: (n = 2) Etc. So, every element of this set is of the form 3 + 4n where n is a whole number. Thus, X = fx : x = 3 + 4n where n 2 Wg : (Read this: X is equal to the set of all x such that x is equal to 3 plus 4n where n is a whole number.) In the previous example, we were successful in writing the set X in set-builder notation. There is one problem though with our nal answer. Notice that the rst element of the sequence, 3; corresponds with n = 0: It would be nice if the rst element corresponded to n = 1: Suppose that you were asked for the 100th term of this sequence. With what we have done so far you would have to realize that the 100th term of this sequence corresponds to n = 99: So, the 100th term is x = 3 + 4 (99) = 399: Let s see if we can x this problem. That is, we want to write X is set-builder notation so that the rst term corresponds to n = 1; the second term to n = 2; etc. Example 16 Continuing the previous example. We know that all terms of this sequence are found by adding some multiple of 4 to the number 3. Let s look at the rst few terms. term 1 = 3 = 3 + 4 0 = 3 + 4 (1 1) = 3 + 4 (n 1) where n = 1 term 2 = 7 = 3 + 4 1 = 3 + 4 (2 1) = 3 + 4 (n 1) where n = 2 term 3 = 11 = 3 + 4 2 = 3 + 4 (3 1) = 3 + 4 (n 1) where n = 3 It appears that this pattern will continue. So, we have that every term of the sequence is of the form 3+4 (n 1) where n is a natural number. Also, this time the value of n is equal to the position of the term in the sequence. That is, n = 1 corresponds to the rst term, n = 100 corresponds to the 100th term, etc. So, X = fx : x = 3 + 4 (n 1) where n 2 Ng : We should simplify our formula. Notice that 3 + 4 (n 1) = 3 + 4n 4 = 4n 1: So, X = fx : x = 4n 1 where n 2 Ng : Now, if we want to determine the 100th term of this sequence, we simply use n = 100 in the formula above to get x = 4 (100) 1 = 399:

Set Theory 13 Set Operations In this section, we will learn how to create new sets from given sets. Union Our rst set operation is the union of sets. De nition 17 Let A and B be sets. The union of A and B; denoted A [ B; is the set containing all elements which are in at least one of the two original sets. In set builder notation we have A [ B = fx : x 2 A and/or x 2 Bg : 1 (1.1) Example 18 Let A = f 1; 1g and let B = f0; 1; 2; 3g : Then A [ B is the set of all elements that are in at least one of the two original sets. So, A [ B = f 1; 0; 1; 2; 3g : Notice that the number 1 is contained in both of the original sets. We do not list it twice in A [ B since we are giving A [ B in roster notation without redundancy. Sometimes we can determine information about sets simply by being told the union of the sets as in the example below. Example 19 Suppose that A; B and C are sets and that A [ B = f1; 2; 3g and A [ C = f1; 2; 3; 5g and B [ C = f1; 2; 3; 5g : Since A [ B does not contain the number 5; we know that the number 5 is not an element of A and is not an element of B: However, 5 is an element of A [ C; so 5 must be an element of A and/or of C: Since 5 is not an element of A; it must be that 5 is an element of C: In this example, we cannot determine much more than this. In fact, here are two possibilities for A; B and C: 1. A = f1; 2g ; B = f2; 3g ; and C = f1; 3; 5g : 2. A = f1g ; B = f2; 3g ; C = f1; 2; 3; 5g : Question: Can you nd another choice of A, B and C that satisfy the conditions of this example, i.e. A [ B = f1; 2; 3g and A [ C = f1; 2; 3; 5g and B [ C = f1; 2; 3; 5g? Intersection De nition 20 Let A and B be sets. The intersection of A and B; denoted A \ B; is the set containing all elements which are in both of the original sets. That is, A\B contains those elements common to both A and B: In set builder notation we have A \ B = fx : x 2 A and x 2 Bg : 2 (1.2) Example 21 Let A = f 1; 1g and let B = f0; 1; 2; 3g : Then A \ B is the set of all elements that are common to the two sets. The only element common to both is the number 1. So, A \ B = f1g : Example 22 Let C = 1; 1 2 ; p 2 and D = N = f1; 2; 3; 4; : : :g : Then C \ D is the set of all elements that are common to the sets. Since there are no elements common to these two sets, we have C \ D = ;: 1 This is read A union B is equal to the set of all x such that x is an element of A and/or x is an element of B: 2 This is read A intersect B is equal to the set of all x such that x is an element of A and x is an element of B:

14 Set Theory Set Di erence De nition 23 Let A and B be sets. The set di erence of A and B; denoted A B; is the set containing all elements which are in A but are not in B: In set builder notation we have A B = fx : x 2 A and x 62 Bg : (1.3) Example 24 Let A = f 1; 1g and B = f0; 1; 2; 3g : Then A B is the set of all elements in A that are not in B: The elements of A are 1 and 1: The only one of these that is in B is 1: So, A B = f 1g : Similarly, B A is the set of all elements in B that are not in A: The elements of B are 0; 1; 2 and 3: The only one of these that is in A is 1: So, B A = f0; 2; 3g : Example 25 Let C = 1; 1 2 ; p 2 and D = N = f1; 2; 3 : : :g : Then C D = 1; 1 2 ; p 2 since none of the elements of C are in D: Similarly, D C = f1; 2; 3; : : :g : Example 26 Let X = f1; 2; 3; 4; 5g and let Y = N = f1; 2; 3 : : :g : Then X Y contains the elements of X that are not in Y: However, every element of X is an element of Y; so there are no elements of X that are not in Y: Thus, X Y = ;: On the other hand, Y X contains the elements of Y that are not in X: So, Y X = f6; 7; 8; : : :g : Complement Our nal set operation is the complement of a set. The complement of a set can be thought of as the set of things needed to complete the set. In order to understand what this means, we have to understand what we mean by complete. For this, we introduce the concept of a universe. The universe is a universal set which all other sets, within a particular discussion, are a subset of. Example 27 Suppose we are discussing the sets A = f1; 1g ; B = f1; 3; 5g and C = f0; 5g : We can not tell from this information what the universe is. However, each of these sets must be a subset of the universe. So, the smallest possible universe is f 1; 0; 1; 3; 5g : On the other hand, it is possible that the universe is Z: Or, the universe could be R: Or, the universe could be fx : x 2 Z and 5 x 5g = f 5; 4; 3; 2; 1; 0; 1; 2; 3; 4; 5g : As you can see from the previous example, the universe cannot be determined uniquely from the given sets. De nition 28 Let A be a set. The complement of A; denoted A; is the set containing all elements of the universe which are not in A: In set builder notation we have A = fx : x 2 U and x 62 Ag : (1.4) Where U denotes the universe. Since it is generally assumed that all elements are in the universe (that s why it s called the universe), this is sometimes written as A = fx : x 62 Ag :

Set Theory 15 Notice that A is simply the set of objects that you need to combine with the set of objects in A to get the universe. That is A [ A = U: Also, by de nition of A; no element can be in both A and A: So, A \ A = ;: Finally, notice that if we change the universe, we change the complement of a set. Example 29 Let A = f1; 2; 3g : We cannot compute A until we are told what the universe is. 1. Suppose U = f1; 2; 3; 4; 5g : Then A contains those elements of f1; 2; 3; 4; 5g that are not in A: So, A = f4; 5g : 2. Suppose U = N: Then A = f4; 5; 6; : : :g : 3. Suppose U = W: Then A = f0; 4; 5; 6; : : :g : 4. Suppose U = Z: Then A = f: : : ; 2; 1; 0; 4; 5; 6; : : :g : Relationships You may have realized by now that there is a strong relationship between set-di erence and complement. In this section we will make this relationship clear. In the previous section, we de ned A = fx : x 2 U and x 62 Ag : According to equation 1.3 in the section on set di erence, the de nition of U A is U A = fx : x 2 U and x 62 Ag : So, A = U A: Now, consider A B: Recall that A B is the set of objects that are in A but are not in B: Also, consider A \ B: This is the set of objects in both A and B: However, if an object is in B; then it is not in B (see the de nition of B): So, if an object is in both A and B; then it is both in A and not in B: So, A \ B contains those objects that are in A and not in B: This is the same as the description of A B: So, A B = A \ B: One advantage of this last relationship is that, to compute A B; you do not need to know the universe. So, you can compute A \ B even without knowing the universe by using the fact that A \ B = A B: Properties In this section we will be studying some of the properties of these set operations. Throughout this course we will be studying properties of operations (e.g. union of sets, addition of integers, multiplication of real numbers, etc.), so we will lay out the general versions of these properties here. Each property is a property of an operation (or two) on a collection of objects. The general statements below are written in terms of a generic operation and set of objects. Commutative Property of Operation of Objects : For any two objects a and b; a ~ b = b ~ a:

16 Set Theory Associative Property of Operation of Objects : For any three objects a; b and c; (a ~ b) ~ c = a ~ (b ~ c) : Distributive Property of Operation 1 over Operation 2 of Objects : For any three objects a; b and c; a ~ 1 (b ~ 2 c) = (a ~ 1 b) ~ 2 (a ~ 1 c) : Closure of Operation of Objects : For any two objects a and b; a ~ b is an object. Using the generic statements above as models, we can state some speci c properties. Note, simply stating a property does not make it true. In each case we will have to ask ourselves whether or not the property is a true property. If a property is true, we will have to make a general argument stating why it is true. This is because each of the properties that we are studying are statements about all objects. So, the property is not true just because the equation works for a few objects, it must work for all of them. On the other hand, suppose we want to argue that one of these properties is false. In that case, we have to show that the equation is not always true. So, we simply need to nd a counterexample; that is, a speci c example of objects for which the prescribed equation is not true. If we can do this, then the equation cannot be true for all objects, so we have shown that the property is false. Example 30 State the commutative property of unions of sets. In this case the operation is union, [; and the objects are sets. So, The commutative property of unions of sets: For any two sets A and B; A[B = B[A: Is this a true property? Suppose A and B are sets. Then A [ B is the set of objects that are in A and/or in B: Clearly, this is the same as the set of objects that are in B and/or in A, that is B [ A: Thus, A [ B = B [ A: So, this is a true property. Example 31 State the commutative property of subtraction of integers. (This is not a set theory example, but we will be studying this type of property later.) In this case the operation is subtraction, ; and the objects are integers. So, The commutative property of subtraction of integers: For any two integers a and b; a b = b a: Is this a true property? No. Let a = 3 and b = 2: Then a and b are integers, but a b = 3 2 = 1 and b a = 2 3 = 1; so a b 6= b a: Therefore, this is a false property. Example 32 State the associative property of intersection of sets. In this case the operation is intersection, \, and the objects are sets. So, The associative property of integersection of sets: For any three sets A; B and C; (A \ B) \ C = A \ (B \ C) : Is this a true property? Yes, for any sets A, B and C; both sides of the equation above are sets containing all elements common to the three original sets. This is a true property. Example 33 State the distributive property of union over set di erence of sets. In this case operation 1 is union, [; operation 2 is set di erence, ; and the objects are sets. The distributive property of union over set di erence of sets: For any three sets A; B and C; A [ (B C) = (A [ B) (A [ C) :

Set Theory 17 Is this a true property? No. Let A = f1; 2; 3g ; B = f2; 3; 4g and C = f3; 4; 5g : Then and A [ (B C) = f1; 2; 3g [ (f2; 3; 4g f3; 4; 5g) = f1; 2; 3g [ f2g = f1; 2; 3g (A [ B) (A [ C) = (f1; 2; 3g [ f2; 3; 4g) (f1; 2; 3g [ f3; 4; 5g) = f1; 2; 3; 4g f1; 2; 3; 4; 5g = ;: So, A [ (B C) 6= (A [ B) (A [ C) : So, this is a false property. Example 34 State the closure property of addition of natural numbers. In this case the operation is addition, +; and the objects are natural numbers. So, Closure of addition of natural numbers: For any two natural numbers a and b; a + b is a natural number. Is this a true property? Yes. The sum of any two natural numbers is a natural number. This is a true property. Example 35 State the closure property of subtraction of natural numbers. In this case the operation is subtraction, ; and the objects are natural numbers. So, Closure of subtraction of natural numbers: For any two natural numbers a and b; a b is a natural number. Is this a true property? No. Let a = 3 and b = 5; then a b = 3 5 = 2 is not a natural number. So, this is a false property. Note that the example above does not show us that closure of subtraction is false. It shows us that closure of subtraction of natural numbers is false. Consider the following example: Example 36 State the closure property of subtraction of integers. In this case the operation is subtraction, ; and the objects are integers. So, Closure of subtraction of integers: For any two integers a and b; a b is an integer. Is this a true property? Yes. If you choose any integer and subtract any other integer from it, the result will be an integer. This is a true property. Venn Diagrams Venn diagrams give us a visual representation of generic sets and set properties. We will use venn diagrams to help us prove (or disprove) some properties of set operations like those studied in the previous section. When discussing two sets, say A and B, we start with the basic venn diagram below

18 Set Theory Where the rectangle represents the universe and the two circles represent the two sets. Notice that there are four regions in this picture. Each region corresponds to a possibility for where an object in the universe lives with respect to A and B: 1. An object may be in neither of the two sets A and B: In this case, the object would be in the region shaded below. 2. An object may be in set A; but not in set B: In this case, the object would be in the region shaded below. 3. An object may be in set B; but not in set A: In this case, the object would be in the region shaded below. 4. Finally, an object may be in both set A and set B: In this case, the object would be in the region shaded below. When discussing three sets, say A; B and C; we simply need to include a third circle to represent the third set. How many regions should we end up with? To determine the answer to this question, let s consider all of the possibilities for where an object in the universe can live with respect to the sets A; B and C: 1. The object may be in none of the sets A; B and C: 2. The object may be in set A; but not in set B and not in set C: 3. The object may be in set B; but not in set A and not in set C:

Set Theory 19 4. The object may be in set C; but not in set A and not in set B: 5. The object may be in sets A and B; but not in set C: 6. The object may be in sets A and C; but not in set B: 7. The object may be in sets B and C; but not in set A: 8. The object may be in all three sets, A; B and C: So, there are eight possibilities. So, the venn diagram should have eight regions. The standard way of representing this is as follows. Exercise 37 Identify which region in the venn diagram above corresponds to each of the eight possibilities listed above. It is very di cult to draw an appropriate venn diagram when four or more sets are being discussed. With four sets, there would need to be 16 regions. Can you determine, as we did above for two sets and for three sets, what the 16 possibilities are? For ve sets, there would have to be 32 regions! So, we usually use venn diagrams to help us when studying situations with two or three sets. Our primary use of venn diagrams will be to prove or disprove properties of set operations. Example 38 Earlier (see page 15) we argued that, for any sets A and B; we have A B = A \ B: Now, let s prove it using venn diagrams. If this statement is true, then the venn diagram representing A B should be identical to the venn diagram for A \ B. Since there are only two sets involved, we will be using the basic two set venn diagram. In the venn diagram for A B; we need to shade all regions containing elements in A B: Since A B contains all elements in A but not in B; the diagram is A B

20 Set Theory Now, we need to nd the venn diagram for A \ B: This will take two steps since there are two operations here, complement and intersection. We start with B: The diagram for B should have all regions shaded containing elements of B; that is all regions containing elements not in B: B Now, A \ B contains all elements common to both A and B: So, we look at the diagram for B and determine which regions shaded there are common to A to get A \ B Notice that the venn diagram for A B is the same as the venn diagram for A \ B: So, for any two sets A and B; we have A B = A \ B: Now, let s use venn diagrams to determine whether a property is true or false. Example 39 Is the associative property of set di erence of sets a true property? To answer this let s rst state the property 3 : For any three sets A; B and C; (A B) C = A (B C) : So, we want to compare the venn diagrams for (A B) C and A (B C) : Let s start with A B: We want objects/regions in A but not in B: 3 Refer to page 16 if you need a reminder on associative property statements. A B

Set Theory 21 So, the venn diagram for (A B) C will have regions shaded that are in A B; but not in C: Now, we need to work on A (B C) : We start with B C: (A B) C B C So, A (B C) will be those regions in A not shaded in the diagram of B C above. A (B C) Since the venn diagrams for (A B) C and A (B C) are not the same, we have that the associative property of set di erence of sets is a false property. When we rst discussed these properties, we decided that, to show that a property was false, we needed to give a counterexample. We can use the venn diagrams determined above to help us come up with a counterexample for the associative property of set di erence of sets. Example 40 Use the information gathered in the previous example to help determine a counterexample to the associative property of set di erence of sets, that is, to determine sets A; B and C such that A (B C) 6= (A B) C:

22 Set Theory If we look at the venn diagrams for A (B C) and (A B) C we see that there are two regions shaded in A (B C) that are not shaded in (A B) C: So, for our counterexample, we need to make sure that there are elements in at least one of these regions. Notice that both of these regions include elements that are in both A and C: So, let s choose sets so that A and C have some common elements. Let s start with A = f1; 2; 3g : (I chose this randomly something simple.) We didn t determine any requirements on B above, so let s stick with something simple here too. Let B = f3; 4g : We need A and C to have common elements, so let s choose C = f1; 4g : Now, let s see if we have really determined a counterexample. And So, A (B C) 6= (A B) C: A (B C) = f1; 2; 3g (f3; 4g f1; 4g) = f1; 2; 3g f3g = f1; 2g : (A B) C = (f1; 2; 3g f3; 4g) f1; 4g = f1; 2g f1; 4g = f2g : Equivalent Sets Recall that two sets are equal if they contain exactly the same elements. So, f1; 2; 3g is equal to f2; 3; 1g since both of these sets contain the elements 1; 2 and 3: While the set f1; 2; 3g is NOT equal to the set fa; b; cg ; these two sets do have something in common. They each have three elements. In fact, these sets can be used to help us understand the concept of three. Ask yourself how you would respond if a child asked you, what does 3 mean? You would probably show him many collections of three objects. You might hold up three ngers, place three pennies on his desk, draw a picture of three apples, etc. That is, you would show the child many sets with three elements each. These sets are called equivalent sets. You would also probably show the child how these collections are related. You might take your three ngers and place one nger on each of the three pennies. In this way you are demonstrating the threeness shared by these two collections. When you do this you are demonstrating a oneto-one correspondence between the two sets. De nition 41 Two sets, A and B; are said to be equivalent, written A B; if there is a one-to-one correspondence between the elements of the two sets. Example 42 Show that f1; 2; 3g is equivalent to fa; b; cg : We need to demonstrate a one-to-one correspondence between these two sets: 1! a 2! b 3! c

Set Theory 23 Note that there will generally be more than one one-to-one correspondence between two equivalent sets. For example, in the example above, we could have chosen 1! c 2! a 3! b instead. We simply have to make sure that we use every element of each set and that no element is corresponded to more than one other element. It is fairly simple, as shown above, to show that two nite sets are equivalent. We simply describe the correspondence completely. What if the sets were in nite? We could never list all of the corresponding elements. Example 43 Let A = f2; 4; 6; : : :g and let B = f5; 8; 11; : : :g : Show A B: An obvious correspondence would be 2! 5 4! 8 6! 11 However, this is an incomplete description. For example, 20 2 A; so 20 must correspond to some element of B: Which one? That is 20!??? We need to come up with a formula to describe this correspondence completely and generically. Each of these sets contains elements forming an arithmetic sequence (see page 11). Using the techniques we developed for writing these types of sets in set-builder notation we get A = fx : x = 2 + 2n where n 2 Wg ; and B = fx : x = 5 + 3n where n 2 Wg : It seems that the generic correspondence should be 2 + 2n! 5 + 3n (where n 2 W): We can check this by plugging in values for n to see if we get the correspondence we started this example with. When n = 0; we get 2! 5: When n = 1; we get 4! 8. Etc. Example 44 Let A = fx : x = 3n + 1 where n 2 Ng and let B = fx : x = 5n + 3 where n 2 Wg : Show A B: Before we do anything else, let s make sure we understand these sets. Let s write each in roster notation. The elements of A are of the form 3n+1 where n is a natural number. When n = 1; 3n+1 = 4: When n = 2; 3n + 1 = 7: When n = 3; 3n + 1 = 10: Etc. So, A = f4; 7; 10; : : :g : Similarly, B = f3; 8; 13; : : :g : As before, we expect our correspondence to be. 4! 3 7! 8 10! 13.

24 Set Theory Based on the initial descriptions of A and B; we might want to use the generic correspondence 3n + 1! 5n + 3: Notice that this does not give us the speci c correspondence we expected. For example, when n = 1 this gives us 4! 8; but we expected 4! 3: Why didn t this work this time. In our previous example, the value of n came from the same set (W) : However, this time n 2 N in A; but n 2 W in B: So, to x this we simply need to write both sets A and B with the same indexing set (i.e. both sets using n 2 N or both sets using n 2 W): Set A is a set whose elements form an arithmetic sequence. Using the technique we learned for writing such a set in set-builder notation, we get A = fx : x = 4 + 3n where n 2 Wg : Now we have A = fx : x = 4 + 3n where n 2 Wg and B = fx : x = 5n + 3 where n 2 Wg : So, the generic correspondence is 4 + 3n! 5n + 3 where n 2 W: Finite vs. In nite Sets We used the terms nite set and in nite set in the previous section. Now, it s time to give a careful de nition of these terms. De nition 45 A set A is said to be nite if 1. A f1; 2; : : : ; ng for some natural number n; or 2. A = ;: Note that A f1; 2; : : : ; ng basically means that the elements of A can be counted o. It says that the elements of A can be corresponded to the numbers 1, 2, 3,..., n: So, there is a 1st element, 2nd element,..., nth element of A: Example 46 Let A = fa; b; cg : Then A is a nite set since, as we showed in example 42 on page 22, A f1; 2; 3g : De nition 47 If a set A is not a nite set, then A is an in nite set. Chapter 1 Exercises 1. List each set in roster notation. (a) A = fx : x = 4n + 5 where n is a natural numberg (b) B = x : x is a whole number and 4:7 x < 23 2 2. List each set in set builder notation.

Set Theory 25 (a) D = f3; 7; 11; 15; : : :g (b) E = f1; 7; 13; 19; 25; : : :g 3. Let S = ffag ; b; fbg ; fa; bgg : Label each of the following as true or false. (a) a 2 S (b) b 2 S (c) a S (d) fag 2 S (e) fag S (f) fbg 2 S (g) fbg S (h) fa; bg 2 S (i) fa; bg S (j) ffagg S (k) fa; fbgg S (l) ffag ; bg S 4. A = f; 4; ~; f~g ; f; 4gg Determine whether each of the following sets is a subset of A; an element of A; both a subset and an element of A or neither a subset nor an element of A: (a) f~g (b) f; 4g (c) f; f~gg (d) f; 4; ~g 5. U = f;; a; fag ; fa; bg ; b; c; d; e; ff gg ; 1; 2; f2gg A = f;; 1; a; c; eg B = f;; ffgg ; a; b; fa; bg ; eg C = f1; 2; f2g ; f;g ; fa; bgg Find each of the following. List each set without redundancy. (a) A [ B (b) A \ C (c) B \ C (d) B [ C (e) C \ A [ B 6. Let A = 1 2 ; p 2; 10 ; B = 1 2 ; 3 2 ; C = ;; and D = f5; 10; 15g : (a) Determine the smallest possible universe for this collection of sets. (b) Determine A (using the universe you found in part a)

26 Set Theory (c) Determine C:(using the universe you found in part a) 7. Let A = fx : x is a natural numberg and let B = fx : x < 10g : Let C = A \ B: Specify C in roster notation. 8. State each of the following set properties. (a) Commutative property for intersection of sets. (b) Associative property for unions of sets. (c) Distributive property for intersection over union of sets. 9. Complete exercise 37 (regarding the three set venn diagram) on page 19 10. Draw a Venn diagram to represent each of the following. For your nal answer draw a diagram with only the appropriate region shaded. (a) A \ B (b) A [ B 11. Use Venn Diagrams in each part to determine whether or not the two sets are equal. (a) (A [ B); A [ B (b) (A [ B); A \ B 12. U = fa; e; f; i; j; f1g ; 2; f2g ; 3; f g ; f;gg A = fa; ff gg ; f2g ; ig ; B = fa; 2; ;; 3; f gg ; C = fe; f; f2g ; f1g ; ;; 2; 5 In each part determine whether or not the two sets are equivalent. Give an explicit 1-1 correspondence for each equivalent pair or justify your answer if the sets are not equivalent. (a) A; B (b) C, A (c) C B; A (d) A [ B; C 13. A = f;g ; 4; fxg ; f g ; 6 3 ; B = f2; x; ;; p; f gg ; and C = 2g n o 5; 6; fyg ; fxg ; ;; x; p 23 4 Determine whether or not each of the following pairs is equivalent and justify each answer. Give a 1-1 correspondence for each equivalent pair. (a) A; B (b) A [ B; C (c) C; A [ B (d) A \ C; B B C 14. Let A = f3; 9; 15; 21; 27; : : :g : Let B = fx : x = 4n 3 where n is a natural numberg : (a) Specify A using set builder notation. (b) Specify B using roster notation. (c) Are sets A and B equivalent? If so, nd a 1-1 correspondence between A and B:

Chapter 2 Arithmetic of the Natural Numbers Development of the Natural Numbers *** oral beginnings historical breakthrough of place value systems need for a place holder later development of 0 as nothing development of N and W as cardinal numbers of a set 27

28 Arithmetic of the Natural Numbers Developing a New Number System (the fen system) *** oral beginnings how many words needed? place value even/odd numbers Activity Fenland Economics I In Fenland, they have four denominations of money. The lowest denomination is blue. Five blues is worth a yellow. Five yellows is worth a green. Five greens is worth a pink. Bank tellers in Fenland are trained to make only one exchange at a time. That is, a teller will make any of the four exchanges listed above, but will not combine them. For example, you can take a green to the bank and the teller will give you ve yellows in exchange for it. The teller WILL NOT give you four yellows and ve blues in exchange for a green. If you want to make such an exchange you will have to do it one exchange at a time. That is, you can exchange your green for ve yellows. You can, then, exchange one of your yellows for ve blues. You should keep track of all exchanges you make as you work through the following activity. 1. You start with 3 greens, 2 yellows and 4 blues. Draw a gure representing this. 2. On payday you receive 2 green, 2 yellow and 2 blue. Using the smallest number of bills possible, describe how much you now have. Describe any exchanges you make along the way to determining the answer. 3. You receive a bill for 3 green, 1 yellow and 3 blue. After paying the bill, how much are you left with. Use the smallest number of bills possible to describe your answer. Describe any exchanges you make along the way to determining the answer.

Arithmetic of the Natural Numbers 29 This Fenland economics activity gives us a nice model for fen numbers. Notice that in the fen number system The lowest place value is the ones. Five ones is a fen. Five fens is a fefen. Five fefens is a fefefen. So, for example, 2 pink, 3 green, 1 blue models the number two fefefen, three fefen, one, that is 2301 f : 1. Above you started with 3 green, 2 yellow and 4 blue. (a) Give the name of the fen number that this represents. (b) Give the fen numeral that this represents. 2. Use your work in part 2 above to complete the following equation 324 f + 222 f = 3. Write the fen equation that your work in part 3 above corresponds to. Addition De nition Consider how you were rst taught to add 3 + 2: You probably did something like this: if I have 3 apples and someone gives me 2 more apples, how many apples will I have? In this example, you are taking a set with 3 objects and a set with 2 objects and nding the size of the union of these two sets. We are going to use this idea to create a careful de nition of addition of whole numbers. De nition 48 Let a and b be whole numbers. To nd the sum, a + b; nd two sets A and B such that 1. # (A) = a; 2. # (B) = b; and 3. A \ B = ;: Then, a + b = # (A [ B) : Example 49 Compute 3 + 5: Let A = f1; 2; 3g and let B = fx; y; z; w; tg : Then # (A) = 3 and # (B) = 5 and A \ B = ;: So, 3 + 5 = # (A [ B) = # (f1; 2; 3; x; y; z; w; tg) = 8: Thus, 3 + 5 = 8:

30 Arithmetic of the Natural Numbers We can use the de nition to prove some properties of addition of whole numbers. For example, we can use it to prove the commutative property of addition of whole numbers. Commutative Property of Addition of Whole Numbers: For any two whole numbers a; and b, a + b = b + a: Proof: Let a and b be whole numbers. Let A and B be sets so that # (A) = a; # (B) = b and A \ B = ;: Then, by de nition, a + b = # (A [ B) : Also, since A \ B = ;; we have B \ A = ;: So, by de nition b + a = # (B [ A) : By the commutative property of unions of sets, we have A [ B = B [ A: So, a + b = # (A [ B) = # (B [ A) = b + a: So, the commutative property of addition of whole numbers is a true property. Algorithm *** look at usual algorithms and need for basic addition table/facts + fen 0 fen 1 fen 2 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 0 fen 0 fen 1 fen 2 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 1 fen 1 fen 2 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 2 fen 2 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 3 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 4 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 10 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 11 fen 11 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 31 fen 12 fen 12 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 31 fen 32 fen 13 fen 13 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 31 fen 32 fen 33 fen 14 fen 14 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 31 fen 32 fen 33 fen 34 fen 20 fen 20 fen 21 fen 22 fen 23 fen 24 fen 30 fen 31 fen 32 fen 33 fen 34 fen 40 fen look at place value/expanded notation as way of building to algorithm

Arithmetic of the Natural Numbers 31 Activity Cuisinaire Rods Colors, from shortest to longest, are: white, red, light green, purple, yellow, dark green, black, brown, blue, orange 1. Use Cuisenaire Rods to show: (a) 2 + 3 = 5 (b) 4 + 2 = 6 (c) 8 + 1 = 9 2. Demonstrate 3 + 5 = 5 + 3 using Cuisenaire rods. 3. Number 2, above, is an example of what property of arithmetic? Subtraction A First Model Take-Away Our rst model of subtraction is usually the take-away model. So, for example, to compute 5 2 we would ask, If I have 5 of something, say cherries, and someone takes away 2 of them, how many cherries do I have left? This is a very good model for subtraction and we will come back to it frequently. However, as we will soon see, there are other models of subtraction to consider.

32 Arithmetic of the Natural Numbers De nition We de ne subtraction in terms of addition. De nition 50 For whole numbers a and b; we say that a solution to a = b + c: b = c if c is the only whole number Example 51 Determine 5 5 2 = 3: 2: Since c = 3 is the only whole number solution to 5 = 2 + c; we have Algorithm *** fen 0 fen 1 fen 2 fen 3 fen 4 fen 10 fen 11 fen 12 fen 13 fen 14 fen 20 fen 0 fen 0 fen 1 fen 1 fen 0 fen 2 fen 2 fen 1 fen 0 fen 3 fen 3 fen 2 fen 1 fen 0 fen 4 fen 4 fen 3 fen 2 fen 1 fen 0 fen 10 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen 11 fen 11 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen 12 fen 12 fen 11 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen 13 fen 13 fen 12 fen 11 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen 14 fen 14 fen 13 fen 12 fen 11 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen 20 fen 20 fen 14 fen 13 fen 12 fen 11 fen 10 fen 4 fen 3 fen 2 fen 1 fen 0 fen Models for Subtraction Subtraction (word) problems can be categorized into four basic models: 1. the take-away model, 2. the missing addend model, 3. the comparison model, and 4. the geometric (or numberline) model. Take-Away Model Take-away model problems generally involve some collection of objects from which a some of the objects are taken or given away. Example 52 Jack has 7 crayons. He give 3 of his crayons to his sister Grace. How many crayons does Jack have left? Missing Addend Model The missing addend model refers to the de nition of subtraction. Remember a b = c means that a = b + c: So, if you are trying to nd a b; then you are trying to nd c: In the equation a = b + c; c is called an addend. If you don t know c; then it would be a missing addend. In using the missing addend model you will generally be computing a b by asking, what do I have to add to b to get a?

Arithmetic of the Natural Numbers 33 Example 53 Robert has 8 toys. He wants to have 12 toys so that he can share with all of his friends. How many more toys does Robert need to get so that he will have enough toys to share with his friends? In missing addend model problems there is usually a goal. In the example above, the goal was 12 toys. Robert needed to know how many toys he needed to add to his 8 toys to get to his goal of 12 toys. That is, he needed to solve 8 +? = 12: This model is frequently used by cashiers. Suppose you are going to the movies and a ticket is $7. You pay with $10. The cashier must determine (unless the cash register tells her) how much change to give you, that is how much more than the ticket price of $7 you gave her. I refer to this as the counting up method of subtraction because the cashier will usually count o dollar bills into your hand saying, 8, 9, 10, and you will end up with $3 in your hand. There is no need to do any subtraction in this case; the cashier only needs to nd the missing addend. 7 +? = 10 I use this method for other (non-word problem) subtraction problems as well. Suppose I want to compute 57 23: I start at 23 and count up until I reach 27, 24, 25, 26, 27. Then, I count up by 10 s until I reach 57, 37, 47, 57. Notice that I counted up by 1 s four times and by 10 s three times, so my answer three 10 s plus four 1 s which is 34. That is, 57 23 = 34: Comparison Model In comparison model problems you are generally comparing two sets of objects. Example 54 Ms. Fidler s class has 25 students while Mr. Hoey s class has 22 students. How many more students does Ms. Fidler have in her class as compared to Mr. Hoey? Note that comparison model problems are frequently written in missing addend language. For example, we could rewrite the last sentence in the example above as How many more students does Mr. Hoey need to get to have as many students as Ms. Fidler does? without changing what the problem is asking. In this case we are asking 22 +? = 25: You can distinguish a comparison model problem from a missing addend model problem by noting that in a comparison problem you will be comparing two distinct collections while in a missing addend model problem you are generally comparing a current collection to some goal. Geometric Model The geometric (or numberline) model is very speci c in that it is used when a numberline is naturally appropriate. For example, if the numbers in the problem represent distance or length. Geometric model problems will fall into one of the previous models as well. What distinguishes them as primarily geometric model problems is the geometric nature of the problem, i.e. distance, length, etc. Example 55 In the morning Ann walks 5 kilometers due east from home to school. After school she walks 2 kilometers west, towards home, and stops at her grandmothers house. How far is the walk home from her grandmothers house?

34 Arithmetic of the Natural Numbers The numberline picture above is a natural representation of the problem and we can easily see that the distance left from grandmothers house to home is 3 kilometers. You must be careful when describing geometric model problems to keep the motion on a straight line. If we leave out the directions in the previous example we run into trouble. Consider the following variation on the previous example. Example 56 In the morning Ann walks 5 km from home to school. After school she walks 2 km and stops at her grandmothers house. How far is the walk home from her grandmothers house? This answer to this problem cannot be determined. Activity Place Value Game 1. As the number cube is rolled, ll in each square with the single digit rolled so as to create the LARGEST possible value. You must enter each set of digits AS IT IS ROLLED. (No fair saving them all up and then lling in the boxes!) 2. As the number cube is rolled, ll in each square with the single digit rolled so as to create the SMALLEST possible value. You must enter each set of digits AS IT IS ROLLED. (No fair saving them all up and then lling in the boxes!) 3. As the number cube is rolled, ll in each square with the single digit rolled so as to create the LARGEST possible SUM. You must enter each set of digits AS IT IS ROLLED. (No fair saving them all up and then lling in the boxes!)

Arithmetic of the Natural Numbers 35 4. As the number cube is rolled, ll in each square with the single digit rolled so as to create the SMALLEST possible POSITIVE DIFFERENCE. You must enter each set of digits AS IT IS ROLLED. (No fair saving them all up and then lling in the boxes!) 5. Would this game be a good classroom activity for kids? Why/Why not? 6. Explain your strategy for each part of this activity. Activity Base 10 Blocks In-Class Activity - Base 10 Blocks: You will be using base 10 blocks for this activity. However, cut up graph paper could be used as well. units rods ats cube Find the Missing Numbers. Give the answer(s) as a numeral and draw a picture of the blocks. Please give all solutions to each question.: 1. I have 5 base-ten blocks. Some are rods and some are units. Their value is less than 20. Who am I? 2. I have 8 base-ten blocks. Some are units and some are rods. Their value is an odd number between 50 and 60. Who am I? 3. I have 4 base-ten blocks. Some are rods and some are units. Their value is between 30 and 40. Who am I?

36 Arithmetic of the Natural Numbers 4. I have 6 base-ten blocks. Some are ats, some are rods and some are units. I am a palindrome. Who am I? Adding and Subtracting with Blocks 1. Draw a sketch which shows how the base ten numeral 234 would be represented using base 10 pieces. 2. Suppose you have 2 ats, 7 rods and 6 units and your partner has 1 at, 4 rods and 9 units. (a) Put all these pieces together and describe (sketch) the exchanges required to keep the same total count but using the smallest number of pieces. Show each exchange. (b) What mathematical calculation does the manipulation in part (a) represent. (c) Explain brie y but clearly how the exchanges in part (a) relate to the mathematical calculation in part (b). 3. Suppose you have 4 ats and 3 units on your desk and you want to remove a count represented by 6 rods and 7 units. (a) Describe (sketch) the exchanges that must take place to accomplish this task. Show each exchange. (b) After removing the 6 rods and 7 units, what pieces are left on the desk? (c) What mathematical calculation does the manipulation in part (a) represent. (d) Explain brie y but clearly how the exchanges in part (a) relate to the mathematical calculation in part (c). Summary (you may discuss these questions in groups, but please don t give group answers; I want to know what YOU think) 1. Give a summary describing what you learned from this activity. 2. Would you use base 10 blocks to teach addition and subtraction to children? Explain your reasons. Converting between Fen and Ten Now that we understand place value well, we are ready to learn how to translate a number given in fen numerals to its base 10 (usual) notation and how to take a number given in standard (base 10) form and give its fen numeral. Fen to Ten Suppose we are given a number written in fen numerals. For example, 32401 fen : To convert this to base 10, we simply have to recall that the fen place value system is based on powers of fen (5 in base 10 notation). That is, from right to left, we have the ones 5 0 position, the fen 5 1 position, the fefen 5 2 position, etc. So, 32401 fen = 3 5 4 + 2 5 3 + 4 5 2 + 0 5 1 + 1 5 0 :

Arithmetic of the Natural Numbers 37 Notice that the left-hand side of this equation is written in fen numerals, but the right-hand side of the equation is written in base ten numerals. So, we have translated from fen to ten. Simplifying, we get Ten to Fen 32401 fen = 3 625 + 2 125 + 4 25 + 0 5 + 1 1 = 1875 + 250 + 100 + 0 + 1 = 2226: To take a number given in standard, base ten, notation and convert it to its base fen notation we simply need to reverse the process above. Notice that, in the example in the previous section, if we write 2226 as a sum of multiples of powers of 5 we get 3 5 4 + 2 5 3 + 4 5 2 + 0 5 1 + 1 5 0 : The coe cients of this sum, 3; 2; 4; 0 and 1; are the digits of the fen numeral 32401 fen : Example 57 Convert the number 3094 to its base fen notation. We need to start by determining the largest power of 5 that can be subtracted from 3094: The powers of 5 are 5 0 = 1 5 1 = 5 5 2 = 25 5 3 = 125 5 4 = 625 5 5 = 3125 So, the largest power of 5 contained in 3094 is 5 4 = 625: We can take out 4 of these since 4 5 4 = 2500: After taking this out we are left with 3094 4 5 4 = 594: Now, we can take out some 5 3 s. We can take out 4 of these since 4 5 3 = 500: After taking this out we are left with 594 500 = 94: Now, we can take out some 5 2 s. We can take out 3 of these since 3 5 2 = 75: After taking this out we are left with 94 75 = 19: Now, we can take out some 5 s. We can take out 3 of these since 3 (5) = 15: After taking this out we are left with 19 15 = 4: So, we are left with 4 (1) ; that is 4 5 0 : Putting all of this together we get 3094 = 4 5 4 + 4 5 3 + 3 5 2 + 3 (5) + 4 5 0 : So, taking the coe cients, we get 3094 = 44334 fen. : Example 58 Convert the number 178 to its base fen notation. Following the process from the previous example we get 178 = 1 5 3 + 2 5 2 + 0 5 1 + 3 5 0 : So, 178 = 1203 fen : Notice that it is important to include the fact that no 5 1 s are removed this time (i.e. 0 5 1 ) so that we do not skip any place value..

38 Arithmetic of the Natural Numbers Multiplication De nition The de nition of multiplication of whole numbers that we will be using is often referred to as multiplication by repeated addition. You may recall have learned originally to multiply this way. For example, to compute 34 you were told to add 4 to itself 3 times. That is 34 = 4+4+4: De nition 59 For any two whole numbers a and b; we de ne a b to be the sum found by adding b to itself a times, that is a b = b + b + {z + } b: a copies of b Example 60 Use the de nition above to compute each of the following. 1. 2 5 = 5 + 5 = 10 2. 5 2 = 2 + 2 + 2 + 2 + 2 = 10 3. 3 0 = 0 + 0 + 0 4. 0 3? In this case we need to add 3 to itself 0 times. This is interpretted to mean 0 3 = 0: Array Model of Multiplication *** relate to area use to show comm, assoc, dist properties of multiplication expanded notation with array model multiplying polynomials with the array model Algorithm *** build up using expanded notation expanded algorithm also use array on these rst two usual algorithm Lattice Algorithm *** Activity Fenland Economics II In Fenland, they have four denominations of money. The lowest denomination is blue.

Arithmetic of the Natural Numbers 39 Five blues is worth a yellow. Five yellows is worth a green. Five greens is worth a pink. Bank tellers in Fenland are trained to make only one exchange at a time. That is, a teller will make any of the four exchanges listed above, but will not combine them. For example, you can take a green to the bank and the teller will give you ve yellows in exchange for it. The teller WILL NOT give you four yellows and ve blues in exchange for a green. If you want to make such an exchange you will have to do it one exchange at a time. That is, you can exchange your green for ve yellows. You can, then, exchange one of your yellows for ve blues. You should keep track of all exchanges you make as you work through the following activity. 1. You start with 3 pink, 1 green, 2 yellow and 2 blue. Draw a gure representing this. 2. You want to distribute this money equally amongst four friends. Complete this process making as few bank transactions as possible. Describe any exchanges you make along the way to determining the answer. 3. How much does each friend receive? Do you have any money leftover? This Fenland economics activity gives us a nice model for fen numbers. Notice that in the fen number system The lowest place value is the ones. Five ones is a fen. Five fens is a fefen. Five fefens is a fefefen. So, for example, 2 pink, 3 green, 1 blue models the number two fefefen, three fefen, one, that is 2301 f : 1. Above you started with 3 pink, 1 green, 2 yellow and 2 blue. (a) Give the name of the fen number that this represents. (b) Give the fen numeral that this represents. 2. Write the fen equation that your work in part 2 above corresponds to. Division (without remainder) De nition Just as subtraction was de ned in terms of addition, division is de ned in terms of multiplication. De nition 61 For whole numbers a and b; we say that a b = c if c is the only whole number solution to a = b c:

40 Arithmetic of the Natural Numbers It is important to note that c must be the only solution as we will see in one of the examples below. Example 62 Determine each of the following possible. If the quotient is unde ned, explain why. 1. 5 0: This is unde ned since the equation 5 = 0 c has no solution. 2. 0 5: 0 5 = 0 since c = 0 is the only solution to 0 = 5 c: 3. 0 0: This is unde ned since the equation 0 = 0 c has more than one solution. (Note: c = 1 and c = 2 are just two solutions since 0 = 0 1 and 0 = 0 2: There are, of course, in nitely many solutions to the equation 0 = 0 c:) 4. 4 3: This is unde ned since the equation 4 = 3 c has no whole number solution. 5. 8 2: 8 2 = 4 since c = 4 is the only solution to 8 = 2 c: As you can see in the examples above, there are two ways for a b to be unde ned. One possibility is that a = b c has no solutions and the other possibility is that a = b c has more than one solution (as with 0 0): Recall that multiplication was de ned as repeated addition. Since division is de ned in terms of multiplication and subtraction is de ned in terms of addition, we might expect division to somehow be related to repeated subtraction. Suppose ab = c: Then a = bc: By the commutative property of multiplication of whole numbers, this is equivalent to a = c b: The repeated addition de nition of multiplication allows us to rewrite this as a = b + b + {z + } b: c copies of b So, if a b = c; then c copies of b can be subtracted from a: Example 63 Use the repeated subtraction concept to compute 24 6: 24 6 = 18 18 6 = 12 12 6 = 6 6 6 = 0: So, four 6 s can be subtracted from 24. Thus, 24 6 = 4:

Arithmetic of the Natural Numbers 41 Models for Division Division (word) problems can be categorized into three basic models: 1. the missing factor model, 2. the division by grouping model, and 3. the division by sharing model. Missing Factor Model The missing factor model refers to the de nition of division. Remember a b = c means that a = b c: In the equation a = b c; c is called a factor. If you don t know c; then it would be a missing factor. In using the missing factor model you will generally be computing a b by asking, what do I have to multiply by b to get a? Example 64 How many hours, at $8 per hour do you have to work to earn $32? Division by Grouping Model The division by grouping model is also referred to as the repeated subtraction model. We discussed the repeated subtraction concept in the previous section. Let s look at an example. Example 65 Ms. Johnson has 12 students in her class. She wants to put the students into groups of 4. How many groups will there be? To see why this model is called repeated subtraction/division by grouping. Consider how Ms. Johnson would make her groups of 4. She would take the 12 students and subtract o one group of 4. 12 4 = 8 Leaving her with 8 students. Then, she would subtract o another group of 4. (12 4) 4 = 4 Leaving her with 4 students which she would put into a nal group of 4.

42 Arithmetic of the Natural Numbers Leaving no students left over. ((12 4) 4) 4 = 0 Division by Sharing Model You may have used the division by sharing model when you did the Fenland Economics II Activity on page 38. This is one for you, one for me, one for you, one for me model that kids use when sharing something such as Halloween candy. Example 66 Ms. Johnson has 12 students in her class. She wants to split them up into four teams. How many students will be on each team? How is this sharing? Think of it as Ms. Johnson sharing her 12 students among the four teams. Division with Remainder De nition We know that 27 6 is unde ned since there is no whole number c such that 27 = 6 c: However, in school we learn to compute this and get 27 6 = 4 R: 3: That is 27 6 is 4 with a remainder of 3. In long division notation this looks like What does this mean? To gure this out let s apply the division by grouping concept to the problem 27 6: In this model we start with 27 and create as many groups of 6 as possible.

Arithmetic of the Natural Numbers 43 We see that we get 4 groups of 6 with 3 left over that can t be put into a group of 6. So, we have 27 = 6 4 + 3: Notice that there was nothing special about 27 and 6 in the discussion above. If you take any numbers a and b, you can create, from a objects, groupings of b objects. You will generally have leftover objects, as we did above. If you end up with q groupings of b objects and r things left over, then you will have that a = bq + r: Notice that you will have fewer than b items left over (that is r < b), since if you had b or more things left over you could pull out another grouping of b objects. This generalized version of division with remainder is called the division algorithm. Theorem 67 (Division Algorithm) Let a and b be whole numbers with b 6= 0: Then there exist unique integers q and r such that a = bq + r and 0 r < b: To nd the q and r guaranteed by the division algorithm you simply perform the division a b: q is the quotient and r is the remainder. That is, a = bq + r; if a b = q R. r: Algorithm *** long division algorithm sca old algorithm standard algorithm short division algorithm Mental Arithmetic and Estimation *** Chapter 2 Exercises 1. Write in the words you would use to say each of the following fen numerals. (a) 241 fen (b) 3; 201 fen (c) 12; 010 fen (d) 20; 310; 414 fen 2. Write the fen numeral for each of the following fen numbers. (a) three fefen one fen two (b) four fefefen two fen one (c) two fen four fefefen three (d) one fefefen, one fen one (e) two fen three

44 Arithmetic of the Natural Numbers 3. In a base fen economy, as you would expect, they use a curious currency. Their paper money is in four denominations. The lowest denomination is the blue. Five blues are worth one yellow, ve yellows in turn are worth one green and ve greens are worth a pink. (a) At your job in Fenland (where they use a base fen economy), you get paid twice a month. Each time you get paid you receive one green, two yellows and three blues. Using the smallest number of bills possible, describe your monthly pay. Describe any exchanges you make along the way to determining the answer. (b) You threw a big New Year s Eve party and now you have no money. Luckily, your boss decides to give everyone a New Year s bonus. She gives you a bonus of two greens, one yellow and one blue. You re very grateful for this, since you still have one more bill to pay from the party. The bill is for two blues and one yellow. After paying the bill, how much do you have left (use the smallest number of bills possible in your answer). Describe any exchanges you make along the way to determining the answer. 4. Give an example of a set which can be used to represent the number 4. 5. Use the de nition (using set theory) of addition to show that 3 + 5 = 8: 6. What property from set theory is used to prove the commutative property of whole number addition. 7. Recall that in the Fenland economy the lowest denomination is blue, ve blues make a yellow, ve yellows make a green and ve greens make a pink. (a) Write the base fen name and the base fen numeral that would represent two blues, three yellows and one pink. (b) Now, you pay a bill for four blues. Give the base fen name and numeral for the remaining money. (do your computations using the money not the numerals) (c) The bill was in error, so you get your money back and now you once again have two blues, three yellows and one pink. You are saving up for a vacation to Ingland. The cost of the vacation is three pinks. Give the base fen name and numeral for the amount of money you still need in order to pay for the vacation. (make your computations using the money not the numerals) (d) Now show the computations you made in parts (b) and (c) using fen numerals. Each part was either a base fen addition or a base fen subtraction problem. Your answer should show one calculation for part (b) and one calculation for part (c) please label which is which. 8. Perform each of the following base fen computations. Show any/all work. You must do all work using base fen numerals. (a) 12 fen + 120 fen (b) 43 fen + 24 fen (c) 302 fen + 44 fen 9. What addition fact does the Cuisenaire Rod picture below represent?

Arithmetic of the Natural Numbers 45 10. Use Cuisenaire Rods to represent each of the following: (a) 4 + 5 = 9 (b) 3 + 3 + 2 = 8 11. Use Cuisenaire Rods and the de nition of subtraction to explain why 8 5 = 3: 12. Perform each of the following base fen computations. Show any/all work. You must do all work using base fen numerals. (a) 32 fen (b) 42 fen (c) 402 fen 11 fen 13 fen 13 fen 13. Use the de nition of subtraction to write each of the following subtraction facts as an equivalent addition fact. (a) 8 4 = 3 (b) 15 9 = 6 (c) 24 fen (d) 331 fen 13 fen = 11 fen 213 fen = 113 fen 14. Use the de nition of subtraction of natural numbers to explain why 3 5 is unde ned. 15. Use the de nition of subtraction of natural numbers to explain why 5 3 = 2: 16. For each of the following word problems determine which model of subtraction the problem best ts into. (a) Megan wants to paint 6 rooms in her new house. So far she has painted 2 of them. How many more rooms does she need to paint? (b) Mary and her brother Ken went trick-or-treating on Halloween. At the end of the night Mary had 53 pieces of candy and Ken had 47 pieces of candy. How many more pieces of candy does Mary have compared with Ken? (c) On Monday, Roberto hiked 11 miles to the lake. On Friday at noon he had hiked 6 miles back down the trail. How much farther does Roberto have to hike to get back to the trailhead? 17. For each of the following models of subtraction, come up with a word problem that ts the model.

46 Arithmetic of the Natural Numbers 18.. (a) comparison model (b) take away model (c) missing addend model (d) geometric model (a) List (in words) any/all exchanges necessary to compute 327 49: (b) Use the standard algorithm (i.e. the one you learned in school) to compute 327 49: (c) Explain where each of the exchanges you described in part a show up in your computation in part b. 19. Convert each of the following base fen numerals into an equivalent base ten numeral. Show any relevant work. (a) 324 fen (b) 12; 003 fen (c) 3; 024; 414 fen 20. Convert each of the following base ten numerals into an equivalent base fen numeral. (a) 200 (b) 117 (c) 20; 346 21. State each of the following properties of whole numbers. (a) Commutative property of addition of whole numbers. (b) Associative property of multiplication of whole numbers. (c) Distributive property of multiplication over addition of whole numbers. 22. Use the de nition of multiplication (as repeated addition) to compute each of the following. Make sure that you show your work so that I can tell that you re using the de nition. 23.. (a) 4 9 (b) 9 4 (c) 2 fen 4 fen (d) 3 fen 10 fen (a) Use the de nition of multiplication (as repeated addition) to compute 15 3: (b) Use the de nition of multiplication (as repeated addition) to compute 3 15: (c) Which was easier to compute?

Arithmetic of the Natural Numbers 47 (d) If you had to make the computation mentally (i.e. in your head without paper and pencil or a calculator) which form of the problem (15 3 or 3 15) would you choose and why? (e) What property of whole numbers tells us that the computations from parts a and b or this problem will yield the same result. 24. Use the array model of multiplication to show/explain why 4 9 = 9 4: 25. Complete the fen multiplication table below: fen 0 fen 1 fen 2 fen 3 fen 4 fen 0 fen 1 fen 2 fen 2 fen 0 fen 3 fen 22 fen 4 fen 4 fen 26. Use the lattice algorithm to complete each of the following computations. (a) 312 42 (b) 312 fen 42 fen 27. In each of the following, determine which property of whole numbers justi es the equality. (a) 4 9 = 9 4 (b) 4 (6 + 2) = 4 6 + 4 2 (c) 3 + 8 = 8 + 3 (d) 12 + (3 + 2) = (12 + 3) + 2 (e) 5 (9 11) = (5 9) 11 (f) 7 3 + 7 8 = 7 (3 + 8) 28. For each of the following subtraction problems (taken from Exploring Mathematics, Grade 2, Scott Foresman and Company), determine which model of subtraction is being used. Then, solve the problem. (a) Amy s book has 124 pages. Steve s book has 133 pages. How many more pages are in Steve s book than Amy s? (b) Jodi s book has 75 pages. On Monday she read 32 pages. How many pages did she have left to read? (c) Lee took 65c/ to the swimming pool. He spent 33c/ for a waterslide ticket. How much did Lee have left?

48 Arithmetic of the Natural Numbers (d) The second graders have 38 more books than the rst graders. The second graders have 87 books. How many books do the rst graders have? 29. Recall that in the Fenland economy the smallest denomination is blue, ve blues is equivalent to one yellow, ve yellows is equivalent to one green and ve greens is equivalent to one pink. (a) You have 1 blue, 4 yellow, 1 green and 2 pink. This money needs to be distributed evenly amongst 4 people. Determine how much each person gets. (give your answer in terms of the money, not as a fen numeral) (b) Show all bank transactions (exchanges) necessary to complete the distribution. That is, go through the process used to determine the answer to part (a) one step at a time. (c) Use fen numerals to describe the division problem that this problem represents. (d) What model of division does this problem t into? 30. For each of the following models of division, come up with a word problem that ts the model. (a) division by sharing model (b) missing factor model (c) division by grouping model 31. Explain why 0 13 = 0: 32. Explain why 13 0 is unde ned. 33. Use the long division algorithm to compute 952 6: 34. Use the sca old algorithm to compute 952 6: 35. Use the short division algorithm to compute 952 6: 36. Place the digits 3; 6; 7 and 8 in the boxes below (using each digit exactly once) to obtain... 37.. (a) the greatest quotient (b) the least quotient (a) Compute 214 fen 3 fen using whichever division algorithm you prefer. (use only fen numerals). (b) Use the result in part (a) to ll in the blanks below: 214 fen = 3 fen + 38. What do you consider to be the best algorithm for multiplication that we have studied. Explain the reasoning behind your choice.

Arithmetic of the Natural Numbers 49 39. Supply reasons for each of the rst three steps given below (525 + 37) + 75 = 525 + (37 + 75) = 525 + (75 + 37) = (525 + 75) + 37 = 600 + 37 = 637 40. Suppose x and y are positive whole numbers. If x is greater than y and you estimate x y by rounding x up and y down will your estimate always be too high or always be too low or could it be either (i.e. sometimes too low, sometimes too high)? Explain. 41. Classroom situations: (a) A student asks if 39 + 41 = 40 + 40; is it true that 39 41 = 40 40: How do you reply? (b) Another student says that he knows that 39 41 6= 40 40 but he found that 39 41 = 40 40 1. He also found that 49 51 = 50 50 1: He wants to know if this pattern continues. How would you respond?

50 Arithmetic of the Natural Numbers

Chapter 3 Number Theory Divisibility Much of number theory centers around the notion of divisibility. De nition 68 Let a and b be integers. We say that b divides a; denoted bja if there is a unique integer q such that a = bq: In this situation we also say 1. b is a factor of a 2. b is a divisor of a 3. a is a multiple of b Recall that when you divide an integer a by another integer b; you get a quotient, q; and a remainder, r: These values are related by the equation a = bq + r: So, if bja; we are getting a = bq; i.e. r = 0: So, bja means that there is no remainder when a is divided by b: That is, that a b is an integer. Even though there is this relationship between the notion of divides and division, this is not the de nition. When working with this concept, it is important to remember that the de nition is as given at the start of this section. Example 69 True or false: 6j24: This is true since 24 = 6 4: Example 70 True or false: 24 is a multiple of 5: This is false since there is no integer q such that 24 = 5q: One way to visualize the notion of divisibility (at least for positive values of a and b) is by using the array model of multiplication. If b divides a; then we have a = bq for some integer q (note that if a and b are positive, then q is also positive). So, there is a b q array using a squares. So, we can determine that b divides a; if we can make an array with b rows using exactly a squares. 51

52 Number Theory Example 71 Use the array model to show that 6j24: The array below has 6 rows and uses 24 squares. In fact, it shows 24 = 6 4: Thus, 6j24: Example 72 Use the array model to show that 24 is not a multiple of 5: If 24 were a multiple of 5, then there would be an array with 5 rows using 24 squares. If I start putting my 24 squares into columns of 5, I will get stuck after I have used 20 of my squares to get and have four boxes left over that can t be used to build another column of 5. Thus, 24 is not a multiple of 5. De nition 73 The natural numbers can be classi ed as prime, composite or unit. A natural number with exactly two positive divisors (itself and 1) is a prime number. A natural number with more than two positive divisors is said to be composite. The natural number 1 is called a unit. Primes There are in nitely many primes Euclid gave the rst known proof that there are in nitely many primes (ca. 300 B.C.). His argument went something like this. Suppose that you have a list of primes, p 1 ; p 2 ; : : : ; p r. Multiply them together and then add 1. Call this number N: So, N = p 1 p 2 p r + 1:. Either N is prime or, by the FTA, it can be factored into primes. If N is prime, then N is a prime that is not on the list because it is larger than any of the primes on the list. If N is not prime, then N must have a prime factor q: Notice that q cannot be on our original list of primes since q divides N evenly, but when N is divided by any of the original primes, p i ; a remainder of 1 is left.

Number Theory 53 Determining whether or not a number is prime To be able to determine whether or not a speci c number is prime, it is helpful to know some relatively small primes. The sieve of Eratosthenes (ca. 200 B.C.) will give us a way to determine these. We will use the sieve of Eratosthenes below to determine all primes between 1 and 200: The technique for using the sieve of Eratosthenes is as follows: 1. Write a list of numbers from 2 to the largest number you want to test for primality (in this case 200). Note: you should cross out the number 1 on the sieve below, since we are only supposed to list the numbers from 2 to 200. 2. Circle the number 2, the rst prime number. 3. Cross o all multiples of 2 from the sieve. 4. The rst number not circled or crossed o remaining in the sieve is a prime number. Circle this number. 5. Cross o all multiples of this number. 6. Repeat steps 4 through 6 until all numbers in the sieve have either been circled or crossed o. 7. The circled numbers are the only primes on the list.

54 Number Theory Now, suppose we want to know whether a number, such as 439 is prime. One option would be to start a new sieve with the numbers from 2 to 439 on it, but that would be a lot of work simply to answer this question. Another option would be to go back to the de nition. If 439 is not prime, then it is composite and that would mean that 439 has at least one other divisor other than 1 and itself, 439: So, we could check for divisors of 439 from among the integers between 2 and 438. That would still be a lot of work. Can we simplify things?

Number Theory 55 First notice that if 439 is not prime, then it must have a prime divisor, so we only need to check for divisors of 439 from among the primes between 2 and 438. To do that we would need to know all of the primes between 2 and 438. We know the primes between 2 and 200, but we don t know the ones between 201 and 438. Can we simplify further? Suppose p is a divisor of 439, other than 1 or 439, then pq = 439 for some integer q: Since p is not equal to 1 or 439, q is also not equal to 1 or 439. So, q 2: So, 1 q 1 439 2 ; so p = q 439 2 = 219:5: So, if 439 is not prime then it must have a prime divisor less than or equal to 219:5: So, we only need to check for divisors of 439 from among the primes between 2 and 219. This is much better since we already know the primes between 2 and 200, but this is still a lot to check. Remember that we found 46 primes between 2 and 200. So, we have more than 46 divisors to check. Can we improve this any more? Suppose p is a divisor of 439. Then pq = 439 for some integer q: Notice that at least one of the two numbers, p and q, must be less than or equal to p 439: To see why, suppose not, so p > p 439 and q > p 439: Then pq > p 439 p 439: But pq = 439 and p 439 p 439 = 439; so this is saying 439 > 439 which is clearly false. Therefore, it must be that one of the two number is less than or equal to p 439: Notice p 439 20:95, so we only need to check for divisors of 439 from among the primes between 2 and 20. That is, check for divisors of 439 from among 2, 3, 5, 7, 11, 13, 17, 19. Let s check: Does 2 divide 439? No. Does 3 divide 439? No. Does 5 divide 439? No. Does 7 divide 439? No. Does 11 divide 439? No. Does 13 divide 439? No. Does 17 divide 439? No. Does 19 divide 439? No. Therefore, since 439 does not have any divisors form among 2, 3, 5, 7, 11, 13, 17, 19, we know that 439 is not composite. Therefore, 439 is prime. Example 74 Determine whether or not 3127 is prime. Based on the work above we only need to check for divisors from among the primes between 2 and p 3127 55:9: So, we need to check to see if any of the following are divisors of 3127 : Does 2 divide 3127? No. Does 3 divide 3127? No. Does 5 divide 3127? No. Does 7 divide 3127? No. Does 11 divide 3127? No. Does 13 divide 3127? No. 2; 3; 5; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 53

56 Number Theory Does 17 divide 3127? No. Does 19 divide 3127? No. Does 23 divide 3127? No. Does 29 divide 3127? No. Does 31 divide 3127? No. Does 37 divide 3127? No. Does 41 divide 3127? No. Does 43 divide 3127? No. Does 47 divide 3127? No. Does 53 divide 3127? Yes, since 3127 = 53 59: Therefore, 3127 has at least 3 divisors: 1, 53 and 3127. So, 3127 is not prime. In the two preceeding examples we have been making use of the following lemma. Lemma 75 If a number n is composite, then there is a prime p such that pjn and p p n: Notice that this lemma does not say that all divisors of n are less than or equal to p n; it simply says that if any prime divisors exist, then at least one of them must be less than or equal to p n: Notice, for example, that in the example above 3127 is a composite and has a prime divisor, 53, which is less than or equal to p 3127: However, 3127 has another prime divisor, 59, which is greater than p 3127: Largest Known Prime As of 17 January 2007, the largest know prime is 2 32582657 1: This prime has 9,808,358 digits. If you want to see whether any larger primes have been found since January 17th, check out www.utm.edu/research/primes/largest.html. This website is a good resource for general information about prime numbers. Fundamental Theorem of Arithmetic Theorem 76 Every natural number (other than the number 1) can be expressed as a product of primes in exactly one way (up to order). This theorem says that any natural number (other than the number 1) can be factored into primes. Furthermore, if two people both factor the number into primes, they will get the same result other than the fact that they might have the primes listed in a di erent order. For example, if two people are factoring the number 12 into primes, one person might get 12 = 3 2 2 while another person gets 12 = 2 3 2: The theorem states that everyone will get two 2 s and one 3 when factoring 12 into primes. How does one nd the prime factorization of a number? The most common method is to use a factor tree. Recall that in a factor tree you start with your number and work downward factoring as you go. You stop when all of the factors are primes. Example 77 Use a factor tree to nd the prime factorization of 1200.

Number Theory 57 So, 2 2 3 2 5 2 5 is the prime factorization of 1200. To make it easier to read, we usually write the prime factorization of a number in prime power representation. When writing a number in prime power representation you put the primes in the factorization in order from smallest to largest and use exponents to simplify, e.g. write 2 3 instead of 2 2 2: Example 78 Determine the prime power representation of 1200. In the previous example we saw that the prime factorization of 1200 contains four 2 s, one 3 and two 5 s. So, the prime power representation of 1200 is 1200 = 2 4 3 5 2 : Number of Divisors *** Lemma 79 A number n written in prime power representation as n = p a 1 1 pa 2 2 pa k k positive divisors. (a 1 + 1) (a 2 + 1) (a k + 1) has Tests for Divisibility Divisibility by 2, 5 and 10 *** Divisibility by 3 and 9 *** Divisibility by 11 *** Divisibility by 2 r *** Divisibility Rules for Composites *** Greatest Common Divisor (GCD) and Least Common Multiple (LCM) De nitions The de nitions of greatest common divisor and least common multiple are very straightforward. We already know what the words greatest, least, common, divisor and multiple mean, so we just

58 Number Theory have to put them together into some de nitions. De nition 80 For two integers a and b; the greatest common divisor of a and b; denoted gcd (a; b) ; is the largest natural number which divides both a and b: De nition 81 For two integers a and b; the least common multiple of a and b; denoted lcm (a; b), is the smallest natural number which is a multiple of both a and b: Set Intersection Method The set intersection method is a precise version of what is usually taught as a rst method for computing gcd s and lcm s. It is nothing more than an application of the de nitions. To use the set intersection method to compute gcd (a; b) you start by nding the set of all divisors of a; denoted D (a) ; and the set of all divisors of b; denoted D (b) : Then D (a) \ D (b) will contain those elements common to both D (a) and D (b) ; that is numbers that are divisors of a and divisors of b; common divisors of a and b: So, the largest element of D (a) \ D (b) will be the greatest common divisor of a and b; gcd (a; b) : Example 82 Use the set intersection method to nd gcd (14; 22) : D (14) = f1; 2; 7; 14g D (22) = f1; 2; 11; 22g D (14) \ D (22) = f1; 2g gcd (14; 22) = 2: Similarly, to nd lcm (a; b) ; we should look at the set of all multiples of a; M (a) ; and the set of all multiples of b; M (b) : Then M (a) \ M (b) will be the set of all common multiples of a and b: So, the smallest element of M (a) \ M (b) will the be the least common multiple of a and b; lcm (a; b) : Example 83 Use the set intersection method to nd lcm (6; 8) : M (6) = f6; 12; 18; 24; 30; 36; 42; 48; : : :g M (8) = f8; 16; 24; 32; 40; 48; 56; 64; : : :g M (6) \ M (8) = f24; 48; : : :g lcm (6; 8) = 24: The set intersection method is a good rst method because it reinforces the de nition of gcd and lcm : However, it is unreasonable to use when the numbers involved are large. (If you don t believe me, try using this method to nd gcd (504; 3675) and lcm (504; 3675) :) Prime Power Representation Method *** Euclidean Algorithm Method *** The Euclidean algorithm is generally the best method for determining gcd (a; b) when the numbers a and b are relatively large. It does not require you to be able to factor the numbers a and b: The following theorem is why the Euclidean algorithm works. Theorem 84 If a = bq + r where a; b; q and r are integers, then gcd (a; b) = gcd (b; r) :

Number Theory 59 Proof. To prove this we will show that D (a) \ D (b) = D (b) \ D (r) ; so that the largest element of D (a) \ D (b) ; i.e. gcd (a; b), is equal to the largest element of D (b) \ D (r) ; i.e. gcd (b; r) : So, we need to show that the common divisors of a and b are precisely the common divisors of b and r: Suppose d is a common divisor of a and b: Then a = dq 1 and b = dq 2 for some integers q 1 and q 2 : So, since a = bq + r; we have dq 1 = (dq 2 ) q + r: Solving for r we get r = dq 1 (dq 2 ) q = d (q 1 q 2 q) : So, d is a divisor of r: We assumed that d was a divisor of b: Thus, d is a divisor of both b and r: So, d is a common divisor of both b and r: Similarly, we can show that a common divisor of b and r must be a common divisor of a and b: Therefore, gcd (a; b) = gcd (b; r) : Activity Interesting Numbers The lyrics of an old song assert that One is the loneliest number. This makes 1 interesting, but all numbers have some property that makes them interesting. 1. 121 I m a perfect square, since I m equal to 11x11, but there is something else interesting about me. What is it? 2. Find three more perfect squares. 3. Find three palindromes. (A palindrome is a number that is unchanged when its digits are reversed.) 4. 64 I m equal to 4x4x4. This means I m a perfect cube. Find something else that is interesting about me. 5. Find three more perfect cubes. 6. What is the smallest multi-digit number that is both a perfect cube and a palindrome? 7. 13 I m prime since my only factors are 1 and 13, but I have another property that not all primes have. What is it? HINT: Reverse my digits. 8. How many palindromes between 100 and 200 are primes? 9. Find three prime numbers such that when their digits are reversed, the result is also a prime number. INTEREST RATING: Prime Number Perfect Cube Perfect Square Sum of Digits Greater than 14 Even Number Odd Number Each Factor 15 points 10 points 7 points 4 points 3 points 2 points 1 point

60 Number Theory Example: 169 Perfect Square (169 = 13 13)... 7 points Sum of Digits Greater Than 14 (1 + 6 + 9 = 16 > 14)... 4 points Odd Number... 2 points Three Factors (1; 13; 169)... 3 points Interest Rating = 7 + 4 + 2 + 3 = 16 points 1. Choose any three-digit number. Find its interest rating using the scale. 2. Can you nd a three-digit number with an interest rating greater than 30 points? Explain. 3. What is the greatest possible interest rating for a three-digit prime number? Explain. 4. Explain why a number greater than 1 that is both a perfect cube and a perfect square would have an interest rating of at least 24 points. 5. Are there any three-digit numbers that are both perfect cubes and perfect squares? If so, what are their interest ratings? 6. Find the highest possible interest rating for a three digit number. Explain. Chapter 3 Exercises 1. Fill in the missing numbers in the following factor tree. 2. Classify each of the following as true or false. Justify your answer using the de nition of divisibility. (a) 11 is a factor of 189 (b) 1001 is a multiple of 13 (c) 6j30 (d) 24 is divisible by 6 (e) 12 is a multiple of 36 (f) 6 divides 2

Number Theory 61 3. Draw an array for each of the following facts: 4.. (a) 4 divides 12 (b) 6 is divisible by 2 (c) 2j8 (d) 20 is a multiple of 5 (e) 3 is a factor of 15 (a) When the U.S. ag had 48 stars, the stars formed a 6 8 rectangular array. In what other rectangular arrays could they have been arranged? (b) How many rectangular arrays of stars could there be if there were only 47 states? 5. Brie y explain why 2 3 3 2 25 3 is not a prime factorization and nd the prime factorization of this number. 6. Produce the smallest positive number that is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11. 7. Find the largest 4-digit number that has exactly three factors. 8. Find a number that has exactly six prime factors. 9. Use a factor tree to nd the prime factorization for each of the following: (a) 504 (b) 2475 (c) 11,250 10. Use the results from problem 9 to determine the number of divisors of each of the following: (a) 504 (b) 2475 (c) 11,250 11. List all divisors of each of the following: (a) 100 (b) 363 (c) 232 (d) 5,929 12. Based on lemma 75, what is the largest prime you must consider to test whether 5669 is prime? 13. The prime numbers 11 and 13 are called twin primes because they di er by 2. Find all twin primes less than 200. (you may nd it helpful to use the Sieve of Eratosthenes)

62 Number Theory 14. Is it possible to nd natural numbers a; b and c such that 2 a 3 b = 5 c? Why or why not? 15. Is 3111 a prime? Prove your answer. 16. Find two whole numbers x and y such that xy = 1; 000; 000 and neither x nor y contains any zeros as digits. 17. Fill in each blank space with a single digit that makes the corresponding statement true. Find all possible answers (a) 3j83_51 (b) 11j8_691 18. If 42jn; what other positive integers necessarily divide n? 19. In order to test for divisibility by 12, one student checked to determine divisibility by 3 and 4, while another checked for divisibility by 2 and 6. Are both students using a correct approach to divisibility by 12? Why or why not? 20. Without using a calculator, test each of the following numbers for divisibility by 2; 3, 4, 5, 6, 8, 9, 10 and 11. (a) 746,988 (b) 15,810 21. When the two missing digits in the following number are replaced, then number is divisible by 99. What is the number? 85 1 22. Classify each of the following as always true or sometimes false: (a) If every digit of a number is divisible by 3, the number itself is divisible by 3. (b) If a number is divisible by 3, then every digit of the number is divisible by 3. (c) If a number is divisible by 6, then it is divisible by 2 and by 3. (d) If a number is divisible by 2 and by 3, then it is divisible by 6. (e) If a number is divisible by 2 and 4, then it is divisible by 8. (f) If a number is divisible by 8, then it is divisible by 2 and by 4. 23. The numbers x and y are divisible by 5. (a) Is xy divisible by 5? Why? (b) Is x y divisible by 5? Why? 24. For each of the following determine all values for the missing digit in the number below so that the number is divisible by 527; 4_2 (a) 2 (b) 3

Number Theory 63 (c) 4 (d) 9 (e) 11 25. Leap years occur in years that are divisible by 4. However, if the year ends in two zeros, in order for the year to be a leap year, it must be divisible by 400. Determine which of the following are leap years: (a) 1775 (b) 1986 (c) 2000 (d) 2024 26. Find the GCD and LCM for each of the following using the intersection-of-sets method: (a) 18 and 10 (b) 24 and 36 (c) 8, 24 and 52 (d) 7 and 9 27. Find the GCD and the LCM for each of the following using the prime factorization method: (a) 132 and 504 (b) 65 and 1690 (c) 900, 96 and 630 (d) 108 and 360 (e) 11 and 19 (f) 625, 750 and 1000 28. Find the GCD for each of the following using the Euclidean algorithm. Then, use the theorem relating GCD s and LCM s to compute the LCM of each pair: (a) 220 and 2924 (b) 14,595 and 10,856 (c) 122,368 and 123,152 29. Bill and Sue both work at night. Bill has every sixth night o and Sue has every eighth night o. If they are both o tonight, how many nights will it be befor ethey are both o again? 30. Jackie spent the same amount of money on cassette tapes that she did on compact discs. If tapes cost $12 and CDs $16, what is the least amount she could have spent on each? 31. By selling cookies at 24c/ each, José made enough money to buy several cans of pop costing 45c/ per can. If he had no money left over after buying the pop, what is the least number of cookies he could have sold?

64 Number Theory 32. At the Party Store, paper plates come in packages of 30, paper cups in packages of 15, and napkins in packages of 20. What is the least number of plates, cups, and napkins that can be purchased so that there is an equal number of each? 33. Find all natural numbers x such that gcd (25; x) = 1 and 1 x 25: 34. Can two numbers have a greatest common multiple? Explain. 35. Is it true that gcd (a; b; c) lcm (a; b; c) = abc? Explain your answer. 36. Suppose that gcd (a; b; c) = 1: Is it necessarily true that gcd (a; b) = 1 and gcd (b; c) = 1? Explain your reasoning. 37. Suppose gcd (a; b) = 2 and gcd (b; c) = 2: Does that always imply that gcd (a; b; c) = 2? Justify your answer. 38. Given that a and b are natural numbers with ab = 1; 515; 025; 512 = 2 3 3 3 7 3 11 2 13 2 and gcd (a; b) = 14; (a) Determine lcm (a; b) without rst nding a and b: (b) Is it possible to determine a and b? Explain.

Chapter 4 The Integers Opposites Up to this point we have been studying the whole numbers. The whole numbers form a useful number system, however, this system has some de ciencies. For example, using only the whole numbers, we cannot solve a simple equation such as 4 + x = 3: This is related to the fact that subtraction is not closed on the whole numbers. Recall (see page 16) that the closure property of subtraction of whole numbers is: for any two whole numbers a and b; a b is a whole number. This is a false property since 3 and 4 are whole numbers, but 3 4 is not a whole number. Once we expand our system of numbers to the integers, we will see that the closure property of subtraction of integers is true. The way to progress from the whole numbers to the integers is by creating additive inverses for each whole number. Let s start with some de nitions. De nition 85 A number a is called an additive identity if a + x = x for every number x: The number 0 has the property that 0 + x = x for every number x; so 0 is an additive identity. So, the whole numbers have an additive identity. De nition 86 For any number x; the number y is called the additive inverse of x if x + y = 0: Notice that the number 3 is a whole number, but there is no whole number y such that 3+y = 0: So, 3 has no additive inverse in the whole numbers. However, the integer 3 is an additive inverse for 3 since 3 + ( 3) = 0: We will refer to the additive inverse of a number as the opposite of that number. So, 3 is the opposite of 3: Here are a few other ways to state the de nition given above for additive inverse/opposite. De nition 87 For any number x; the number y is called the opposite of x; if x + y = 0: De nition 88 The opposite of a number x is the number y that when added to x gives a sum of 0. 65

66 The Integers De nition 89 For any number x; the equation x + y = 0 tells us that y is the opposite of x: The notation that we will use for the opposite of a number x is x: Note that this should be read opposite of x, NOT as negative x as the opposite of a number need not be negative. For example, since 3 + 3 = 0; we have that 3 is the opposite of 3: So, ( 3) = 3 (read this as the opposite of negative 3 is 3, or as the opposite of the opposite of 3 is 3 ). It is permissible to read 3 as negative 3 since it truly is a negative number. It may also be read as opposite of 3, since it is also the opposite of 3. The set of integers (remember the notation for the set of integers is Z) can be partitioned into three subsets: 1. the set of positive integers (this is just the set of natural numbers, N) 2. the set of negative integers (these are the opposites of the natural numbers) 3. the set containing the number 0. It is important to note here that 0 is neither positive nor negative. Using the notation x for the opposite of x; we can restate the de nitions above as follows. De nition 90 The opposite of a number x is the number x such that x + ( x) = 0: De nition 91 x + y = 0 () y = to the opposite of x: ) x (read this as x plus y is equal to 0 if and only if y is equal Representations of the Integers Most of the common representations of integers involve using two-sided objects. One side represents positive and the other, negative. Colored counters are sold at teacher supply stores and are either disk or bean shaped with one side red (representing negative) and the other side white or yellow (representing positive). You can achieve the same e ect by using pennies (heads for positive, tails for negative). You will get some practice with these manipulatives as a representation for the integers in the activity following this section. A similar representation is the charged particle representation. This is essentially the same idea as the counters, except that you do not use a hands-on manipulative. Instead you use positive particles, drawn as, and negative particles, drawn as. An example of how to use these is as follows. Example 92 What integer is represented by the following collection of charged particles?

The Integers 67 Each pair of one positive particle and one negative particle contributes nothing to the overall value (since a negative particle is the opposite of a positive particle and a number plus its opposite is zero). So, after pairing o positive/negative pairs, we are left with so there are two negative particles left. Thus, this collection of charged particles represents 2: The idea of charged particles works well if the students are also studying atoms in their science class. For example, in the previous example, I could have restated the question as, What is the charge of the ion containing two protons and four electrons? Activity Colored Counters PURPOSE: Investigate a concrete model for integers and use it to discover algorithms for integer addition. MATERIALS: A paper cup (optional), 6 pennies, a game marker for each player and a game board (a number line showing the integer values from 10 to 10). First A Game The Rules: