Mathematics 13th Edition, Smith Notes Korey Nishimoto Math Department, Kapiolani Community College October 9 2017
Expanded Introduction to Mathematical Reasoning Page 3 Contents Contents Nature of Logic.................................... 5 Deductive Reasoning............................. 5 3.1.1 Conjunction and Disjunction 5 3.1.2 Negation 6 3.1.3 Homework 3.1 8 Truth Tables and the Conditional....................... 10 3.2.1 The Conditional 10 3.2.2 Converse, Inverse, and Contrapositive 11 3.2.3 Truth Tables 12 3.2.4 Homework 3.2 14 Operators and Laws of Logic......................... 17 3.3.1 The Bi-Conditional 17 3.3.2 Operations (De Morgan s Law) 17 3.3.3 Homework 3.3 19 The Nature of Proof.............................. 21 3.4.1 Direct Proof 21 3.4.2 Proof By Contrapositive 21 3.4.3 Proof By Contradiction 22 3.4.4 Homework 3.4 23 The Nature of Sets.................................. 25 Sets, Subsets, and Venn Diagrams....................... 25 2.1.1 How to Write Sets 25 2.1.2 Venn Diagrams and the Universal and Empty Set 26 2.1.3 Subsets and Proper Subsets 27 2.1.4 How many Subsets are there? 29 2.1.5 vs { } 29 2.1.6 Homework 2.1 30 Operations with Sets............................. 33 2.2.1 Intersections and Unions 33 2.2.2 Cardinality 34 2.2.3 Test Your Knowledge 35 2.2.4 Homework 2.2 37 Application of Sets.............................. 40 2.3.1 Given Numerals and a Picture 40 2.3.2 Draw Venn Diagrams with Appropriate Data 41 2.3.3 Homework 2.3 46 Infinite and Finite Sets............................ 49 2.4.1 Countable and Uncountable 49 2.4.2 Homework 2.4 51 The Nature of Counting............................... 52 Permutations and Counting.......................... 52 12.1.1 Fundamental Counting Principle 52 12.1.2 Permutations 53 12.1.3 Distinguishable Permutations 54 12.1.4 Homework 12.1 56 Combinations................................. 59 12.2.1 Combinations Vs Permutations 59 12.2.2 Binomial Theorem 60 12.2.3 Counting Flow Chart 62 12.2.4 Homework 12.2 63 Counting Without Counting.......................... 65 12.3.1 Random Counting Problems 65
Expanded Introduction to Mathematical Reasoning Page 4 Contents 12.3.2 Homework 12.3 68 The Nature of Probability............................. 70 Introduction to Probability.......................... 70 13.1.1 Theoretical Probability 70 13.1.2 Empirical Probability 73 13.1.3 Homework 13.1 74 Mathematical Expectations.......................... 77 13.2.1 Expected Value 77 13.2.2 Homework 13.2 79 Probability Models.............................. 81 13.3.1 Complement Probability 81 13.3.2 Conditional Probabilities 82 13.3.3 Homework 13.3 84 Calculating Probabilities........................... 87 13.4.1 Probabilities Using Counting Principles 87 13.4.2 Or Probabilities 88 13.4.3 And Probabilities 90 13.4.4 General Probability Instructions 92 13.4.5 Homework 13.4 93 The Nature of Statistics............................... 96 Frequency Distributions............................ 96 14.1.1 Stem and Leaf Plot 97 14.1.2 Bar and Line Graph 97 14.1.3 Pie Graph 99 14.1.4 Combine it All 99 14.1.5 Homework 14.1 103 Descriptive Statistics............................. 106 14.2.1 Measures of Central Tendency 106 14.2.2 Box Plots 107 14.2.3 Homework 14.2 112 The Normal Distribution........................... 115 14.3.1 Determining And Using The Normal Distribution 115 14.3.2 Standard Scores and Percentiles 117 14.3.3 Normal Distribution Instructions 119 14.3.4 Homework 14.3 121 The Nature of Financial Mathematics...................... 124 Interest.................................... 124 11.1.1 Growth and Decay Factor 124 11.1.2 Types of Interest 124 11.1.3 Different Uses of Interest 127 11.1.4 Homework 11.1 129 Month Payments, Loans and Savings..................... 132 11.2.1 Monthly Payments For Savings 132 11.2.2 Monthly Payments For Loans 133 11.2.3 Homework 11.2 136 Financial Project............................... 140
Nature of Logic 3 Nature of Logic Expanded Introduction to Mathematical Reasoning Page 5 3.1 Deductive Reasoning Definition 3.1.1: A statement is a declarative sentence that is either true or false. It cannot be both true and false. Ex: 3.1.1: Some examples of statements and sentences that are not statements. 1) I have a car. 2) I hate you! 3) Hawaii is located on the moon. 4) How are you today? 5) Chickens and cats are animals. 6) This statement is false. Notice number 5 in example 3.1.1 uses the word and. We will now discuss the connecting words of two or more statements. 3.1.1 Conjunction and Disjunction Note: Remembering truth values is difficult if you try to remember the entire truth table. Instead remember the special case for each type of statement discussed and you will have less to remember. Conjunctions and disjunctions are a very important part of logic. They allow us to create more common and useful statements. We will first discuss the compound statement conjunction, then disjunction. Definition 3.1.2: The word and is called a conjunction and is used to combine statements. Let p and q be statements, then p and q is written as p q. The truth table below gives us an understanding of when this compound statement is true (T) or false (F). p q p q T T T T F F F T F F F F Notice that the only time that the compound conjunction is true is when both the left and right hand sides are true. It is easier to remember this rule than each line separately. All you need to do is check if both sides are true. If it is not, the entire conjunctions is automatically false. We will see some examples further down in example 3.1.3. Ex: 3.1.2: Your parents tell you they will give you a car and $1,000,000 if you get straight A s. You get straight A s. If your parents only give you a car, or only $1,000,000, did they lie to you? The answer is yes. Regardless of how happy you are to get a car or $1,000,000 they still lied. The only time that they would not have lied is if they had given you both. It is also clear that they lied if they gave you neither. Definition 3.1.3: The word or is called a disjunction and is used to combine statements. Let p and q be statements, then p or q is written p q. Similarly the truth table will help us understand disjunctions. p q p q T T T T F T F T T F F F
Nature of Logic In each statement we should determine the type of statement which is made. For example 3.1.3 number 7, (p r) (p q) is an or statement or a disjunction. The main connector is highlighted in red. What is the main connector for 8)? Expanded Introduction to Mathematical Reasoning Page 6 Notice in this case the only way a disjunction is false is when both the left and right hand side are false. Same as example 3.1.2, but with statement, If you get straight A s they will give you a car or $1,000,000. Now the only time that your parents would have lied to you is if they gave you neither. One thing to note, in mathematical logic, or is inclusive. This means that we may have one the other or both, to make the statement true. Because of this it is easiest to remember when disjunctions are false. You only need to check to see if both mini statements are false. If they are not then the statement is automatically true. Ex: 3.1.3: Determine the truth value of the given statements. Assume that p and r are true, while q is false. 1) p q T F T 3) r (p q) T (T F ) T T T 2) p q T F F 4) r p T T T 5) (p q) r (T F ) T F T T 7) (p r) (p q) (T T ) (T F ) T F T 6) p q r T F T F 8) (p r) (q (r q)) (T T ) (F (T F )) T (F T ) T T T 3.1.2 Negation In example 3.1.4, 3 we keep the statement p q and negate the entire statement instead of the common error of negating each part separately. We commonly use words like; not, don t, false, and not true. These are all words that give us a negation. Consider the statement I like to eat apples, the negation could be I do not like to eat apples or It is not true that I like to eat apples. The negation of statements including compound statements using disjunctions and conjunctions is very common in our everyday speech. Definition 3.1.4: The negation of a statement changes the truth value of that statement. It is written as. It should be clear that the negation of a negation gives back the original. This is because the truth value switches two times. T F T or F T F. Ex: 3.1.4: Let p represent the statement Nursing informatics is a growing field and let q represent Critical care will always be in demand. Translate the following. 1. p q Nursing informatics is not a growing field and critical care will always be in demand. 2. (p q) It is not the case (It is not true) that nursing informatics is a growing field or critical care will always be in demand. 3. (p q) It is not the case (It is not true) that nursing informatics is a growing field and critical care will always be in demand.
Nature of Logic Notice that in each case we work similar to the acronym PEMDAS that you learned previously. This time there are no exponents, multiplication, division, addition, or subtraction. However you do work from the inside of the parenthesis out, by evaluating each step one at a time. Expanded Introduction to Mathematical Reasoning Page 7 Now that we have seen an example of how to convert the english statements into its negations, lets consider the symbolic version. We want to be able to take a compound statement where we know the truth value of each mini-statement, and apply the negation in the correct order. This is similar to the example 3.1.4, except we will plug in T and F, then apply the negations as necessary. Ex: 3.1.5: Let p represent a false statement and let q represent a true statement. Find the truth value of the given compound statement. 1) ( p) ( F ) T F 3) (p q) (F T ) (F F ) F T 2) (p q) (F T ) T F 4) ( p q) ( F T ) (T F ) F T 5) [ p ( q p)] [ F ( T F )] [T (F F )] [T F ] T F 6) [( p q) q] [( F T ) T ] [(T F ) F ] [T F ] F T
Expanded Introduction to Mathematical Reasoning Page 8 Problem 1. Answer the following questions. 3.1.3 Homework 3.1 1) What is the special case for an or statement? What is every other case? 2) What is the special case for an and statement? What is every other case? 3) What is a negation? 4) What is a statement? 5) Write the truth table for a conjunction. 6) Write the truth table for a disjunction. Problem 2. Answer the following questions. Justify your answer. 1) If q is false, what must be the truth value of the statement (p q) q? 2) If q is true, what must be the truth value of the statement q (q p)?
Expanded Introduction to Mathematical Reasoning Page 9 Problem 3. Let p represent a false statement and let q represent a true statement. Find the truth value of the given compound statement. 1) (p q) 2) ( p q) 3) [ p ( q p)] 4) [( p q) q] Problem 4. Let p represent a true statement, and let q and r represent false statements. Find the truth value of the given compound statements. 1) [ q (r p)] 2) (p q) (p q) Problem 5. True or False. Give an explanation. 1) A statement p q is only false when both p and q are false. 2) A statement p q is only true when both p and q are true. 3) The negation of a statement p is always false. 4) A double negation will always give you the original statement back.
Nature of Logic Expanded Introduction to Mathematical Reasoning Page 10 3.2 Truth Tables and the Conditional We have already seen some truth tables in the previous section and we know that truth tables help us understand the truth value of a given compound statement given any combination of truth values of the mini statements. In this section we will learn how to construct these truth tables along with a new connecting word. 3.2.1 The Conditional Let us first discuss the new connecting logical symbol so that we may use it in our construction of truth tables. Definition 3.2.1: An if-then statement is called a conditional statement. Let p and q be statements. Then the conditional if p, then q is written as p = q. The first statement which is attached to the if is called the antecedent / hypothesis, and the statement attached to the then or after = is called the consequence / conclusion. Ex: 3.2.1: Some examples of the conditional. 1. If Hawaii wasn t so expensive, the island would be completely full. 2. If I don t exercise, then I will gain weight. The conditional is a little weird with respects to the truth value. I will explain after the truth table below. p q p = q T T T T F F F T T F F T Notice that the only time that a conditional statement is false is when T = F. Many people think that if the first statement is false, then the conditional statement must also be false. This is the example I like to think about that helped me understand. Ex: 3.2.2: If it rains today, then the road will be wet. Let p be it rains today, and q be the road will be wet. Here are the cases. 1) T = T It rained today so the road is wet. 3) F = T It didn t rain today but the road is wet. 2) T = F It rained today but the road is not wet. 4) F = F It didn t rain today so the road is not wet. Which one of these doesn t make sense? I think it is pretty clear that 1 and 4 make sense. Most people will agree that 2 also makes sense. The debate usually happens in 3. The question is, it didn t rain so is it possible that the road is wet? Yes. Someone could have spilled water, snow could have melted, or a fire hydrant could have been broken. This statement can happen. You may also consider the statement, If you win the lottery, then you will have a million dollars. One thing to note, conditional statements do not give use cause and effect. An example of this may be If a unicorn is born, then I will have a new iphone or If the sun rises, then monkeys are mammals. Both of these statements are true. The first statement is true since the antecedent is false. This automatically makes the conditional true. The second conditional is true since both the antecedent and the consequence are true. I think it should be clear that these two statements have to cause and effect relation ship. The conditional statement is one of the most commonly used statements and thus has some special versions.
Nature of Logic Note: This procedure works for any conditional statement. If you follow the rules, then the converse, inverse, and contrapositive is easy to find. The converse of the contrapositive is the inverse. The converse says to take the left and the right of the contrapositive and flip them. So q = p converts to q = p which is the inverse. Expanded Introduction to Mathematical Reasoning Page 11 3.2.2 Converse, Inverse, and Contrapositive This section gives people a lot of trouble, and is actually used incorrectly a lot because people believe they are using it correctly. We will discuss this after we state the converse, inverse, and contrapositive. Definition 3.2.2: Given the conditional p = q, we define: 1. Converse: q = p The converse says to take the conditional s left hand side and put it on the left, then take the right hand side and put in on the left. This is the converse. 2. Inverse: p = q The inverse says to take the conditional s left hand side and put negation on it, then take the right hand side and put a negation on it. This is the inverse. 3. Contrapositive: q = p The Contrapositive says to take the conditional s left hand side and put it on the left, then negate it. Then take the right hand side and put in on the left and negate it. This is the contrapositive. Ex: 3.2.3: Lets find the converse, inverse, and contrapositive statements of the following conditionals. 1. p = (q p) (a) The converse says to flip the the left and the right hand sides of the original. (q p) = p (b) The inverse says to negate both the left and the right hand side of the original. p = (q p) (c) The contrapositive says to flip the left and the right hand sides of the original, then negate both of them. 2. (p = q) = (p q) (q p) = p (a) The converse says to flip the left and the right hand sides of the original. (p q) = (p = q) (b) The inverse says to negate both the left and the right hand side of the original. (p = q) = (p q) (c) The contrapositive says to flip the left and the right hand sides of the original, then negate both of them. 3. (p q) = (p q) (p q) = (p = q) (a) The converse says to flip the left and the right hand sides of the original. (p q) = (p q)
Nature of Logic Expanded Introduction to Mathematical Reasoning Page 12 (b) The inverse says to negate both the left and the right hand side of the original. (p q) = (p q) (p q) = (p q) (c) The contrapositive says to flip the left and the right hand sides of the original, then negate both of them. (p q) = (p q) (p q) = (p q) Which of these are the same? Let the class discuss with each other and use examples. Many people believe that the inverse is the same as the conditional. Lets reconsider the statement If it rains today then the road will be wet. Can we infer that If it does not rain, then the road will not be wet. In order to show which of these are the same we will need to construct a truth table. 3.2.3 Truth Tables We have seen truth tables before, and we know that the truth table gives us the truth value of a given statement for all possible combinations of truth values of the mini-statements. Let s discuss the actual method to construct these tables. Instructions 1. Determine the amount of rows needed by calculating 2 n, where n is the number of unique statements. If we have statements p and q, then there are 2 unique statements, and there will be 2 2 = 4 rows. Whereas if there are statements p, q, and r, then there will be 2 3 = 8 rows. 2. Draw the table with the given amount of rows, with columns for each mini statement and the possibly large compound statement. 3. Fill in the truth value for the mini statements. The first column will have the top half T and the bottom half F. Now take the amount of T and divide by 2. This is the amount of T in the second column before alternating to F, then alternating back to T until the entire column is full. Continue this process until all mini statement s columns are full. 4. Now evaluate the truth table and circle the column of your final answer. Construct a truth table for the following. Ex: 3.2.4: Construct a truth table for the following. 1. For the converse, inverse, and contrapositive of the conditional p = q. p q p = q q = p p = q T T T T T T T T F T T F T T F T F F F T T F T T T F F T F T T T F F T F F F T F F F T F F T F T F T T F
Nature of Logic Expanded Introduction to Mathematical Reasoning Page 13 q = p F T T F T T F F F T F T T T F T F T T F Notice that the conditional and the contrapositive have the same truth table, while the inverse and the converse have the same truth table. This means that the end value of the compound statement is the same, or the blues are the same. This tells us that the converse and inverse as well as the conditional and the contrapositive are equivalent statements. This is the only way to determine if two statements are equivalents. You cannot determine equivalence using 1 case of the truth table. All the different cases must be the same in order for the statements to be equivalent. 2. (p = q) = ( p q) p q (p = q) = ( p q) T T T T T T F T T T T F T F F T F T F F F T F T T T T F T T F F F T F T T F T F This is an example of a special compound statement called a Tautology. This is where the compound statement is true regardless of the truth value of the mini statements. Let s now combine everything we learned using conditional, disjunctions, and conjunctions to determine the truth value of the statement with given truth values. Ex: 3.2.5: Determine the truth value of the statement when p is false, q and r is true 1) (p q) = r (F T ) = T T = T T 3) (q p) (p = r) (T F ) (F = T ) F T T 5) (p = q) = (q = r) (F = T ) = (T = T ) T = T T 2) (p r) = (r q) (F T ) = (T T ) T = T T 4) (p = r) (q p) (F = T ) (T F ) T F F 6) (p q r) = (p [q = r]) (F T T ) = (F [T = T ]) F = [F T ] F = F T
Expanded Introduction to Mathematical Reasoning Page 14 Problem 1. True or False. Provide explanation. 3.2.4 Homework 3.2 1) Let p and q be statements and consider the conditional p = q. The contrapositive of the converse statement is the inverse of p = q. 2) Given that p is true and q is false, the conditional p = q is true. 3) Given that p is false and q is false, the conditional p = q is true. 4) 6-9=3 = 8+2=10. 5) If 3+1 7, then 8 8. 6) If 8+7 15, then 3+5=8. Problem 2. Find the truth value of the given statement, given that p is true and q and r are false. 1) [p = (p = q)] ( q r) 2) [(p = q) (p = r)] = (p r) 3) [(p = r) = (q = r)] 4) [ (p r) (q p)] (q r)
Expanded Introduction to Mathematical Reasoning Page 15 Problem 3. Write the converse, inverse, and contrapositive statements for the following conditional statements. 1) 6-9=3 = 8+2=10. 2) If 3+1 7, then 8 8. 3) (p = q) = (p = q) 4) p = q Problem 4. Give a real life example of a conditional statement not used in class. Then convert the condition statement you created to its converse, inverse and contrapositive statements. Your answer should be in English.
Expanded Introduction to Mathematical Reasoning Page 16 Problem 5. Create a truth table for the following compound statements. Label in the following order, p, q, r. 1) ( q = p) = q 2) (p q) = (p q) 3) [( p = q) = (p q)] 4) [ (p q) = (q p)] [(r = q) ( p = r)]
Nature of Logic 3.3 Operators and Laws of Logic Expanded Introduction to Mathematical Reasoning Page 17 Sometimes it is useful to be able to do operations in logic. We may want to simplify before finding the truth value. Simplify does exactly what is sounds like, it simplifies the expression. In this section we will discuss the bi-conditional as well as operations in logic. 3.3.1 The Bi-Conditional We will first discuss the related logical statement the bi-conditional. This is exactly as it sounds 2-conditionals. Definition 3.3.1: An if and only if statement, also written as iff, is called a bi-conditional statement. Let p and q be statements. Then the bi-conditional is written p q. It is a little easier to understand this as Meaning, both must imply each other. (p = q) (q = p). Lets think about this for a second. The statements must imply each other, and conditional statements are false only when T = F. So we get truth table p q p q T T T T F F F T F F F T In this section p. 98, the book says that there is a difference between and = as well as and. They say that the single lined arrow is used of the logic symbol while the double lined arrow is used for logical equivalence. For us, we will always use = and, and if we want to say logically equivalent, then we must first show logical equivalence using a truth table, and then just simply state they are logically equivalent. Lets do this using a rule of logic called De Morgan s law. Theorem 3.3.1: 3.3.2 Operations (De Morgan s Law) De Morgan s law gives us a way to algebraically manipulate the negation of a conjunction or disjunction. (p q) p q (p q) p q We will prove the first part of De Morgan s law p q (p q) p q T T F T T T F T F F T T F F T T F F T F T F F T F F T T T F F F T F F T F F F T F T T F The truth tables are the same, hence (p q) p q Next lets discuss the negation of conditional and bi-conditional statements. Theorem 3.3.2: The negation of a conditional and bi-conditional are given as, (p = q) p q,
Expanded Introduction to Mathematical Reasoning Page 18 Nature of Logic and (p q) [(p = q) (q = p)] (p = q) (q = p) (p q) (q p). Ex: 3.3.1: Use the rules above to find the simplified negation of the following statements. 1) p (q = r) [p (q = r)] p (q = r) p q r 3) (p = q) = (r = p) [(p = q) = (r = p)] (p = q) (r = p) (p = q) (r p) 5) p q (p q) [(p = q) (q = p)] (p = q) (q = p) (p q) (q p) 2) (p = q) (r = p) [(p = q) (r = p)] (p = q) (r = p) (p q) (r p) 4) (p r) (q [r p]) [(p r) (q [r p])] (p r) (q [r p]) ( p r) ( q [r p]) ( p r) ( q r p]) 6) (p = q) (r = p) [(p q) (r p)] [(p q) (r p)] [(r p) (p q)] [(p q) ( r p)] [(r p) ( p q)]
Expanded Introduction to Mathematical Reasoning Page 19 3.3.3 Homework 3.3 Problem 1. Find the simplified negation of each statement. If applicable, use De Morgan s laws. 1) (p = q) = (p r) 2) (m n) (k j) 3) (p r) = (p q) 4) [(p q) = (q r)] = (p = q) Problem 2. Create the truth table for the following statements. 1) (p = q) (q p) 2) [ (p q) ([p = q] p)]
Expanded Introduction to Mathematical Reasoning Page 20 Problem 3. True or False. Must provide explanation. 1) The negation of a conjunction is equivalent to a disjunction. 2) The negation of the contrapositive q = p is the converse q = p. 3) Let p and q be true. Then p = q is true and p q is true. This means that they are logically equivalent. 4) The Bi-conditional statement (p = q) ( p q) is a tautology.
Nature of Logic 3.4 The Nature of Proof Expanded Introduction to Mathematical Reasoning Page 21 In this section we will be discussing different methods that are commonly used in mathematics to prove statements in mathematics. We will first begin with direct reasoning or direct proof. 3.4.1 Direct Proof This uses a straight reasoning of the conditional statement, without any modifications. We must first assume the antecedent (Why is this?) before we start. And use the antecedents information to try to conclude the concequence. Ex: 3.4.1: Prove that the sum of 2 even numbers is even. (Did you know that every even number can be written as 2k, where k is an integer. Integers={... 3, 2, 1, 0, 1, 2, 3...}.) This can be rewritten as a conditional as such if we add two even numbers, then the sum is odd. Now first assume that the antecedent (we add two even numbers) is true. This is the same as 2a + 2b where a and b are integers. Factor a 2 to get 2(a + b). Since a + b is an integer we are of the form 2 times an integer which makes the sum even. 3.4.2 Proof By Contrapositive The next type of proof is called indirect reasoning/contrapositive proof. This uses the contrapositive statement q = p. We do this because sometimes the negation of q has more information than p itself. Lets do two examples like this. Ex: 3.4.2: Prove the following using a contrapositive proof. 1. Prove that if you multiply two integers together and get an even number, then at least one must be even. Did you also know that if a number is not even then it is odd and is written as 2k+1 again where k is an integer. We want to prove the contrapositive and the contrapositive give us, If none of the numbers are even, then when you multiply them together the result is not even/odd. Now we know about even number so assume the antecedent of the contrapositive is true (None of the numbers are even). Then they must be of the form 2a + 1 and 2b + 1 where a and b are integers. Multiplying them together we get (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1. Similarly as in example 3.4.1, ab is an integer and adding three integers is an integer, hence we have 2 times an integer plus 1. This is odd or not even. 2. If x 2 is even, then x is even. Contrapositive: If x is not even, then x 2 is not even. Assume x is not even, hence x = 2a + 1 where a is an integer. This gives us x 2 = (2a + 1) 2 = 4a 2 + 4a + 1 = 2(2a 2 + 2a) + 1. Similarly in ex. 3.4.2 part 1, this is of the form 2 times an integer plus 1, hence not even or odd.
Nature of Logic Expanded Introduction to Mathematical Reasoning Page 22 This works because p = q q = p, shown in 3.2, example 3.2.4. 3.4.3 Proof By Contradiction Lastly we will discuss the negation/proof by contradiction. This is not in the book, but it is one of the fundamental proof methods in math. The negation of p = q is p q. If we show that p q is false, then what must p = q be? It is true because the negation is false. If we negate a true statement then we get a false statement. Our goal is to try to find a contradiction. Ex: 3.4.3: Prove by contradiction. 1. Let n be an integer. If n 2 is odd, then n is odd. The negation give us n 2 is odd and n is even. We can assume both of these to come up with a contradiction. Since n is even n = 2a where a is an integer. Hence n 2 = (2a) 2 = 4a 2 = 2(2a 2 ). This is of the form of 2 times an integer which makes this even. However we first assumed that n 2 is odd. These two conflict and give us a contradiction. 2. Prove if a and b are integers, then a 2 4b 2. The negation gives us a and b are integers and a 2 4b = 2. Notice if we bring b to the right hand side we get, a 2 = 2 + 4b = 2(1 + 2b). This again is the form of an even number so a 2 is even. We proved in example 3.4.2 part 2, that if a 2 is even then so is a. Thus a = 2c where c is an integer. Substituting back we get (2c) 2 4b = 2. We get 4c 2 4b = 2 which gives us 4(c 2 b) = 2. Now divide both sides by to to get 2(c 2 b) = 1. Notice the left hand side is of the form 2 times an integer, which makes it even, but 1 is not even. This is a contradiction.
Expanded Introduction to Mathematical Reasoning Page 23 Problem 1. Prove by contradiction. 1. Let n be an integer. If n is odd, then n 2 is odd. 3.4.4 Homework 3.4 2. Let n be an integer. If n is even, then n 2 is even. Problem 2. Give a contrapositive proof. 1. Let n be an integer. If 3n + 1 is even, then n is odd.
Expanded Introduction to Mathematical Reasoning Page 24 bonus Let m and n be integers. If n 2 + y 2 is even, then n + m is even. Problem 3. Give a direct proof for the following. 1. Prove that the sum of two odd numbers is even. 2. Prove that the sum of three odd numbers is odd.
Sets 2 The Nature of Sets Expanded Introduction to Mathematical Reasoning Page 25 Note: When learning sets, the key is to remember the definition and how to read them. The reason why people cannot do math is because they cannot read math. Practice reading the symbols in a way that you understand it. If you understand it a different way than is being taugh, discuss it with you teacher to ensure that it is correct. 2.1 Sets, Subsets, and Venn Diagrams In this section, we will have a brief introduction to a mathematics called set theory. Definition 2.1.1: A set is a collection of distinct objects called elements. Definition 2.1.2: The complement of a set is the set of all elements not in the set. This is denoted a A where A is the set and A (your book uses Ā) is the complement of set A. When writing a set, mathematicians have certain conventions. For example, a set is always denoted by a capital letter, an element is usually denoted by a lower case letter, and there is a difference between and =, which we will discuss latter. 2.1.1 How to Write Sets First lets discuss some sets using the different methods of writing sets. 1. Listing method (Your book calls this roster method): Ex: 2.1.1: Let try to construct the set of integers between and including 5 and 10 (x Z and 5 x 10). Can we see this includes numbers 5,6,7,8,9, and 10? To write this in set notation all we have to do is give the set a name, call the set of integers between and including 5 and 10 A. Then write A = {5, 6, 7, 8, 9, 10}. The listing method is usually used when we are writing either finite amounts of elements, or there is a pattern to the infinite number of elements. Ex: 2.1.2: x 10. Let B be the numbers larger or equal to 10. B = {10, 11, 12,...}. Lets look back at A. What are the elements of A? They are 5,6,7,8,9, and 10. To write this symbolically, we write 5 A. This is read as 5 is in set A. This is a very common notation and we have already used it once to describe the integers. Here are some more common sets in math. (a) The set of integers=z = {..., 3, 2, 1, 0, 1, 2, 3,...} (b) The set of natural numbers=n = {1, 2, 3, 4, 5,...} (c) The set of whole numbers=w = {0, 1, 2, 3, 4, 5,...} (d) The set of rational numbers=q = { a b a, b Z and b 0} (e) The set of irrational numbers=q = {x x Q} (f) The set of real numbers=r = {x x is on a number line}. Notice that d, e, and f use a different notation which we will now discuss. 2. Set Builder Notation:
Sets Ex: 2.1.3: Expanded Introduction to Mathematical Reasoning Page 26 Let is write A above in set builder notation. A = {x x Z and 5 x 10}. This reads A is equal to the set of x such that x is an integer and is between or including 5 and 10. This type of set is usually used when the set is to cumbersome to write as a list. The difficulty of this type of set is being able to read the symbols. Math is a language and you must be able to read it. Two sets are equal when they have the same elements. Order does not matter. This means A = {1, 2, 3} and B = {3, 1, 2} are the same and A = B. 2.1.2 Venn Diagrams and the Universal and Empty Set Sets are difficult to understand without a picture. The representation of a set as a picture is called a Venn Diagram. Before we can draw these we need two more definition. Definition 2.1.3: The universal set is the set of all possible elements in the given situation. It is denoted as U. 1. For the common sets above U = R. 2. We did an experiment testing a new drug. Does it make sense that the universal set is all people? No. It is all people in the experiment. Note: The empty set is not the same as { }. The empty set is a set which contains nothing (No Elements). However the set { } has one element in it, namely the empty set. Remember to read the set notation {} as the set containing. 3. Three people in this class got A s on the exam. What is the Universal set? All the people in this class. Definition 2.1.4: The empty set is the set that contains no elements. It is denoted as. This is why you cannot write 0 as or {}. Ex: 2.1.4: Let us draw a Venn Diagram for the previous set A = {5, 6, 7, 8, 9, 10} U=R 5,6,7,8,9,10 This tells us that the blue circle only has the elements 5,6,7,8,9, and 10 in it. while the inside the box but outside of the blue circle has the rest of R. One thing to note, Venn diagrams do not need to be circles. Most people use circles as a convention, but it is not necessary. You may have used the shape of a horse to represent set A, and the answer would still be correct provided that 5, 6, 7, 8, 9, and 10 are in A and you label U.
Sets Expanded Introduction to Mathematical Reasoning Page 27 2.1.3 Subsets and Proper Subsets The last thing to discuss in this section is subsets and proper subsets. Definition 2.1.5: A subset B of A is define as such. Every element in B is also in A. This is written as B A and is read B is a subset of A. Definition 2.1.6: A proper subset B of A is define as such. Every element in B is also in A, but A B. We will use the symbol and is read as in. This is written as B A and is read B is a proper subset of A. You can think of this as B is inside of A. Definition 2.1.7: Two sets A and B are disjoint if they share no elements in common. That is for every x A, x B, and for every x B, x A. The only difference between subset and proper subset is that subset allows the possibility of equality where as proper subset does not. When in doubt subset will always work assuming that B is contained in A. Notice that in definition 2.1.7 we use. This is not the same as =. We have shown before that two sets are equal when they have the exact same elements. x A, says that the number x is an element of A, not that x and A are the same. Let us now draw some Venn diagrams which show the different cases that can happen. 1. B and A have nothing in common or are disjoint. U A B 2. B is a proper subset of A or, B A. U A B
Sets Expanded Introduction to Mathematical Reasoning Page 28 3. A = B, which can also be written as A B and B A. U B A One thing to note is the definition of the empty set and subset. The empty set is the set which contains no elements, while the subset definition states a set is a subset of a second set if all of its elements are contained in that second set. What can we sat about the empty set, and its relationship to the subset definition? The empty set is a subset of any set. All its elements (which are none) are contained in any other possible set. Going back to the common math sets. Which of these sets are subsets of each other? Give me some examples. Irrationals Real numbers Rational numbers Integers Whole numbers Natural numbers
Sets Expanded Introduction to Mathematical Reasoning Page 29 2.1.4 How many Subsets are there? Let me pose a question. How many subsets are there of the set A = {5, 6, 7, 8, 9, 10}? There happen to be 64 subsets in total. Lets demonstrate this using a couple slightly smaller sets. Ex: 2.1.5: How many subsets are there for the following sets? 1) A = {1, 2, 3} 2) A = {cat, dog} {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}, {cat}, {dog}, {cat, dog}, There are 8 subsets in total. There are 4 subsets in total. 3) There is 1 subset. 4) A = {2, 4, 6, 8} {2},{4},{6},{8}, {2,4},{2,6},{2,8},{4,6},{4,8},{6,8}, {2,4,6},{2,4,8},{2,6,8},{4,6,8}, {2,4,6,8}, There are 16 subsets in total. Can we see a pattern? Notice that we originaly started with 1 subset when the set was the empty set. Then we moved to 4 when the set had 2 elements, 8 when the set had 3 elements, and lastly 16 when the set had 4 elements. What is the pattern here? Theorem 2.1.1: The total number of subsets is 2 n where n is the number of elements in the set. This will help us know if there is a missing subset. Many times we forget the two simplest ones, the empty set and the entire set. Based on this how many proper subsets do we have? Proper subset is a set contained in another, but is not equal to it. So we have 2 n 1 proper subsets. There is only one subset which is equal. 2.1.5 vs { } An interesting thing to consider is the difference between and { }. What are the elements of and what are the elements of { }? The empty set, or contains no element, while { } contains one element, namely. This is a little confusing, but { } is a set which contains a set as its elements. You can think of this as a group of groups. For examples A = { Dogs, Cats, Mammals, Bugs, Fish}. Each of these categories has a lot of different elements. For example; golden retriever, labradors, chihuahua, pit-bull, and shiba inu, are all different types of dogs. This makes A a set of sets.
Expanded Introduction to Mathematical Reasoning Page 30 2.1.6 Homework 2.1 Problem 1. Insert or in each blank so that the resulting statement is true. 1. {2, 3, 100, 55} {x x is an even integer.} 2. {7, 33, 1 2 } Q 3. {x x is an odd letter in the alphabet.} {x x a type of elephant.} 4. {Boston, Orlando, Honolulu} The set of states. Problem 2. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Find the compliment of each set. 1) A = {2, 4, 6, 8, 10} 2) A = {10, 8, 6, 9, 7, 3, 5, 2, 4, 1} 3) A = {1, 3, 4, 5, 8, 1, 4, 5} 4) A = Problem 3. Draw one venn diagram with all of the following properties. U is the universal set, A B, C B but no elements of C are in A, and no elements of a set D is in A, B, or C.
Expanded Introduction to Mathematical Reasoning Page 31 Problem 4. Find the number of subsets and the number of proper subsets of each set. For 1, also write all the different subsets. 1) {a, b, c, d} 2) The set containing the days of the week. 3) {x x is an odd integer between -4 and 6.} 4) {x x is a whole number less than 4} Problem 5. Draw one venn diagram with all of the following properties. U is the universal set, A B C, A D,and D B and D C Problem 6. True or False. Provide explanation. 1. A is never a subset of A. True False 2. If A B, then A B. True False 3. If A B and B C, then A C True False
Expanded Introduction to Mathematical Reasoning Page 32 Problem 7. 1. Give two real life examples not done in class of sets A and B so that A B. 2. Draw a Venn Diagram for both. 3. Justify in words that the examples you created have the property A B.
Sets 2.2 Operations with Sets Expanded Introduction to Mathematical Reasoning Page 33 In this section we will discuss different types of operations involving sets. This includes intersections (and), union (or), and cardinality (count). We have already discussed and as well as or in the logic section. This is the exact same thing. 2.2.1 Intersections and Unions Note: You can read A B as the elements or things, in both the sets A and B. As a picture, this reads as the overlap of the sets A and B. First lets discuss intersections. Definition 2.2.1: The intersection of two sets A and B is the set of elements that are in both A and B. It is written as A B. The Venn diagram is draw as U A A B B Note: You can read A B as the elements or things, in set A, set B, or in both A and B (In one, the other, or both). As a picture, this reads as the circle A, circle B, or the overlap of A and B. Definition 2.2.2: The union of two sets A and B is the set of elements that are in either A or B. Remember, or is inclusive. This means one the other or both. It is written as A B. The Venn diagram is draw as U A A B B A B=blue For the union, notice that A B includes A, B, and A B. This is the same as saying in A, B, or in both. The union and intersection are similar to the logic version. We had a rule for negations of and and or statements called De Morgan s law. This still holds for sets. Theorem 2.2.1: Let A and B be sets. We have the following properties. (A B) = A B and (A B) = A B.
Sets Ex: 2.2.1: Find the following sets. Expanded Introduction to Mathematical Reasoning Page 34 1) {x, y, z, a, b, c} {xy, yz, ab, bc} This is empty or the. Just because there is an x in both does not mean that x is an element of both. Look carefully in the second set. The element is actually xy not x. 3) { cat, dog, chicken} { dog, ant, bird} A B = {cat, dog, chicken, ant, bird} Don t forget two things. One we want the elements in one, the other, or both, and two, writing elements twice does nothing, so we do not need to write dog again. 2) { 2, 3, 5} { 2, 2, 1 5} They both share 2 and 5. Hence A B = { 2, 5} 4) (Q Q ) (Q Q ) = Q Q = Q Q This reads as in the rationals, in the irrationals, or in both. What is every rational number combined with every irrational number? This is the entire number line. This A B = R. 2.2.2 Cardinality The last thing that we need to discuss is cardinality. Definition 2.2.3: Cardinality is the number of elements in the set. For example let A = {2, 4, 6}. The cardinality of A of n(a) = 3. Your book uses A, but I think that this is confusing since most people know as the absolute value sign. This, n(a), reads as n of A, can be thought of, n, or number of elements of A. Theorem 2.2.2: The cardinality of the union of two sets A and B is If A and B are disjoint, then n(a B) = n(a) + n(b) n(a B). n(a B) = n(a) + n(b). This is because n(a B) =, hence n(a B) = 0. We get this formula because of the way we count sets. Think about this in terms of a Venn Diagram. U A A B B
Sets Expanded Introduction to Mathematical Reasoning Page 35 Why is A B maroon colored? Its because the blue part and the red part overlap each other. You can think of the shading as the counting. That is n(a)=blue shade and n(b)=red shade. If this is the case, notice we have counted A B twice. Once with the ble shade and once with the red shade. This is the reason why we need to take away (subtract) the overlap A B once. Lets do the following examples where the sets are given and we must find the cardinality of each of the following. Ex: 2.2.2: Let A = {x 0 x 10 and x Z}, B = {2, 4, 6, 8}, and C = {3, 9, 27}. Solve for the following. 1) n(a) =11 3) n(c) =3 5) n(a C) = n(a) + n(c) n(a C) = 11 + 3 2 =12 7) n(a B) = n(a) + n(b) n(a B) = 11 + 4 11 =4 This makes sense as well since B A, A B = B 2) n(b) =4 4) n(a B) = n(a) + n(b) n(a B) = 11 + 4 4 = 11 Since B is a subset of A, n(a B) = n(a). 6) n(c B) = n(c) + n(b) n(c B) = 3 + 4 0 =7 8) n(b C) = n(b) + n(c) n(b C) = 4 + 3 7 =0 This makes sense because B and C are disjoint. Thus their intersection is empty. In this example we wil be given the cardinality of certain sets. We must find the cardinality of the following. Ex: 2.2.3: Let n(a) = 17, n(b) = 13, n(c) = 12, n(a B) = 10, and n(b C) = 15. Find the following. 1) n(a B) = n(a) + n(b) n(a B) = 17 + 13 10 = 20 2) n(b C) = n(b) + n(c) n(b C) = 13 + 12 15 = 10 2.2.3 Test Your Knowledge Ex: 2.2.4: Are the following sometimes true, always true, or false? 1. A B A B A B is some times a subset of A B. This happens when A = B. We then have A B=A A = A and A B = A A = A. Thus A A. 2. A B A B This is always true by definition 2.2.2 of A B. A B is the set of all elements
Sets Expanded Introduction to Mathematical Reasoning Page 36 in A, B, or both (A B). 3. n(a B) = n(b A). We know that unions and intersections are commutative. That is A B = B A. Hence n(a B) = n(b A). Now that we have a understanding of cardinality, lets draw a Venn diagram. Ex: 2.2.5: Draw a Venn diagram with the following properties. A B C, A B, and A C. Also shade the following. (A B) C. U B A C
Expanded Introduction to Mathematical Reasoning Page 37 2.2.4 Homework 2.2 Problem 1. Match each term in Group 1 to the appropriate description in group 2. 1 2 1) The complement of A. A. The set of elements that are in A or in B or in both A and B. 2) The union of A and B. B. The set of elements common to both A and B. 3) The intersection of A and B. C. The set of elements in the universal set that are not in A. Problem 2. Use the listing method to describe the following sets. U = {a, b, c, d, e, f, g}, X = {a, c, e, g}, Y = {a, b, c}, Z = {b, c, d, e, f}. 1) X Y 2) Y Z 3) X Y 4) X (Y Z) 5) Y (X Z) 6) (Y Z ) X 7) (X Y ) Z 8) (Z X ) Y 9) (Y X ) Z 10) (X Y ) (Y X )
Expanded Introduction to Mathematical Reasoning Page 38 Problem 3. Let A and B be two sets. Is the following statements always true or sometimes true. Give an explanation. 1) A (A B) 2) A (A B) 3) (A B) A 4) (A B) A 5) n(a B) = n(a) + n(b) 6) n(a B) = n(a) + n(b) n(a B) Problem 4. Draw one venn diagram that represents the following operations. A B, A C and C B =. Shade in (A B ) C.
Expanded Introduction to Mathematical Reasoning Page 39 Problem 5. Give an example of two real life sets, not used in class, A and B such that A B and A B and B A. (Read as not equal to. Anytime there is a line through a symbol, read it as not a. What is?) 1. Draw a Venn diagram and label. 2. Justify that your example represents the properties asked in the question. 3. What is A B as a sentence using your real life examples.
Sets 2.3 Application of Sets Expanded Introduction to Mathematical Reasoning Page 40 Application of sets is a common idea which we have used everyday for almost all our lives. We discuss things like how many people work full time and go to school, or how many native Hawaiian students are in a S.T.E.M degree. It is easiest to jump straight into this topic with examples. Ex: 2.3.1: 2.3.1 Given Numerals and a Picture Use the numerals representing cardinalities in the Venn diagrams to give the cardinality of each set specified. 8 U A 0 5 B 2 1) A B n(a B)=5 3) A B n(a B )=0 2) A B n(a B)=7 4) A B n(a B)=2 5) A B n(a B )=8 Now lets try an example which has three sets instead of two. The general idea of how to find the cardinality of sets is still the same. Ex: 2.3.2: Use the given data to find the cardinality of the given sets. 8 U A 8 3 1 4 0 B 2 C 2 1) A B C n(a B C) = 1 2) A B C n(a B C ) = 3
Sets 3) A B C n(a B C) = 4 Expanded Introduction to Mathematical Reasoning Page 41 4) A B C n(a B C) = 0 5) A B C n(a B C) = 2 7) A B C n(a B C ) = 2 6) A B C n(a B C ) = 8 8) A B C n(a B C ) = 8 As previously stated, the most important part of this is the ability to read the problem. For example, in example 2.3.2 part 3, we want to to the number of elements that are in A and C, but outside of B. This is the most difficult thing to do in math, but the most important. 2.3.2 Draw Venn Diagrams with Appropriate Data We originaly started with the Venn diagram and evaluated the cardinality of each given set. Now we will go backwards. We will be given the cardinality of each set, then draw the Venn Diagram. Ex: 2.3.3: Draw a Venn diagram and use the given information to fill in the number of elements in each region. n(a) = 57, n(a B) = 35, n(a B) = 81, n(a B C) = 15, n(a C) = 21, n(b C) = 25, n(c) = 49, n(b ) = 52 12 U A 16 20 15 6 10 B 14 C 18 Ex: 2.3.4: Find the value of n(b) if n(a) = 20, n(a B) = 6 and n(a B) = 30. Now draw a Venn diagram for this data. n(a B) = n(a) + n(b) n(a B) = n(b) = n(a B) n(a) + n(a B), so n(b) = 30 20 + 6 = 16.
Sets Expanded Introduction to Mathematical Reasoning Page 42 U A 20-6=14 6 B 16-6=10 Ex: 2.3.5: A middle school counselor, attempting to correlate school performance with leisure interests, found the following information for a group of students. 1) 34 had seen Despicable Me 2) 29 had seen Epic 3) 26 had seen Turbo 4) 16 had seen Despicable Me and Epic 5) 12 had seen Despicable Me and Turbo 6) 10 had seen Epic and Turbo 7) 4 had seen all three of these films 8) 5 had seen none of the three films 1. How many students have seen Turbo only? First we find how many people have seen all three. There are 4 people who saw all three and this is given by (7). We also know that n(e T ) = 10 so the number of people who have seen Epic and Turbo but not Despicable me is 6. Do this again for Despicable me and Epic to get that 8 people have seen Despicable me and Turbo, but not Epic. As a Venn diagram we have U D E 4 8 6 T Notice to get the number of people who have seen Turbo, we add 8+4+6+x = 26 by (3). This tells us that the number of people who have seen Turbo only is 8. We can also think of this as n(turbo only)=n(t ) n(d T ) n(e T ) + n(t D E) = 26 12 10 + 4 = 8. 2. How many have seen exactly two of the films? We already have solve two of the parts we need in 1). We only need to find the number of people who have seen Despicable me and Epic, but not Turbo.
Sets Expanded Introduction to Mathematical Reasoning Page 43 Using the same strategy as in 1) we get 16-4=12. In order to get the number of people who have seen exactly two films we need, n(despicable me and Epic only)+n(despicable me and Turbo only)+n(turbo and Epic only) 3. How many students were surveyed? So far we have = 12 + 8 + 6 = 26. U D 12 4 8 6 E T 8 We can solve for the number of people who have seen Despicable me only and Turbo only the same way we solved for the number of people who saw Turbo only in 1). This gives us n(despicable me only)=34-8-4-12=10 and n(epic only)=29-4-6-12=7. The only thing left is to calculate the number of people who saw none. But this is already given by (8). We add all these numbers up to get 8 + 4 + 6 + 12 + 8 + 10 + 7 + 5 = 60. 4. Draw a Venn diagram. 5 U D 10 12 4 8 6 E 7 T 8 Ex: 2.3.6: A survey was conducted among 75 patients admitted to a hospital cardiac unit during a two-week period. Let B= The set of patients with high blood pressure, C= The set of patients with high cholesterol levels, and S= The set of patients who some cigarettes. The survey produced the following data. 1) n(b) = 47 2) n(b S) = 33 3) n(c) = 46 4) n(b C) = 31 5) n(s) = 52 6) n(b C S) = 21
Sets Expanded Introduction to Mathematical Reasoning Page 44 7) n[(b C) (B S) (C S)] = 51 Find the number of these patients who, 1. Had either high blood pressure or high cholesterol levels, but not both. By theorem 2.2.2, n(b C) = n(b) + n(c) n(b C). However we don t want the number of people with high blood pressure or high cholesterol levels. We want n(b C But not both). This says to take away the people with both, hence n(b C But not both) = n(b C) n(b C) = n(b) + n(c) n(b C) n(b C) = 47 + 46 2(31) = 31. 2. Had fewer than two of the indications listed. This statement asks us to find the number of people that have only one or no indications. To make this problem easier, we must first find the only missing intersection n(c S). We will use 7 to do this. As a Venn diagram, 7) give us the following shaded area has 51 people in it. U B C S This is n[(b C) (B S) (C S)] = n(b S)+n(B C)+n(S C) 2n(B S C)=33 + 31 + n(s C) 2(21). Hence 51=22+n(S C) = n(s C) = 29. Using this and the method used in the second explanation of part a) in the example above, n(b only) = n(b) n(b C) n(b S)+n(B S C) = 47 31 33+21 = 4 n(c only) = n(c) n(c B) n(c S)+n(C B C) = 46 31 29+21 = 7 and n(s only) = n(s) n(s C) n(s B)+n(S C C) = 52 29 33+21 = 11 Lastly we need to find the number of people with no indications. We can do this by the following. n(number of people with no indications) = Total number of particpants [n(b) + n(c) + n(s) n(b S) n(c S) n(b C)] = 75 [47 + 46 + 52 33 29 31] = 2. Adding these all up we get the number of people that had fewer than two of the indications listed is 24. 3. Were smokers but had neither high blood pressure nor high cholesterol levels. This question is asking for the number of people who were smokers only. This is given by n(s) n(b S) n(s C) + n(b C S) = 52 33 29 + 21 = 11. 4. Did not have exactly two of the indications listed. This is easier if you solve for the number of people that have exactly two and then deduct that number from the total number of participants. n(b and C only) = n(b C) n(b C S) = 31 21 = 10
Sets Expanded Introduction to Mathematical Reasoning Page 45 n(b and S only) = n(b S) n(b C S) = 33 21 = 12 n(s and C only) = n(s C) n(b C S) = 29 21 = 8. Hence the number of people that did not have exactly two of the indications listed, is 75-(10+12+8)=45. 5. Draw a Venn diagram 2 U B 4 10 21 12 8 C 7 S 11
Expanded Introduction to Mathematical Reasoning Page 46 2.3.3 Homework 2.3 Problem 1. At a southern university, half of the 48 mathematics majors were receiving federal financial aid. 1) 5 had Pell Grants 2) 14 participated in the College Work Study Program 3) 4 had TOPS scholarships 4) 2 had TOPS scholarships and participated in Work Study 5) Those with Pell grants had no other federal aid. Create a Venn diagram and answer how many of the 48 math majors had: 1) No federal aid? 2) More than one of these three forms of aid? 3) Federal aid other than these three forms? 4) A TOPS scholarship or Work Study? 5) Exactly one of these three forms of aid. 6) No more than one of these three forms of aid?
Expanded Introduction to Mathematical Reasoning Page 47 Problem 2. 140 U.S. adults were surveyed. Let A= The set of respondents who believe in astrology, R= The set of respondents who believe on reincarnation, and Y = The set of respondents who believe in the spirituality of yoga. The survey revealed the following information: 1) n(a) = 35 2) n(r) = 36 3) n(y ) = 32 4) n(a R) = 19 5) n(r Y ) = 8 6) n(a Y ) = 10 7) n(a R Y ) = 6 Create a Venn diagram and answer how many of the respondents believe in: 1) Astrology but not reincarnation? 2) At least one of these three things? 3) Reincarnation but neither of the others? 4) Exactly two of these three things? 5) None of the three?
Expanded Introduction to Mathematical Reasoning Page 48 Problem 3. True or False. Give explanation. 1. [(A B) (B C)] = (A B ) (B C ) 2. If A is disjoint from B and B is disjoint from C then A is disjoint from C. Problem 4. Use the numerals representing cardinalities in the Venn diagram to give the cardinality of each set specified. 1) A B C = 2) A B C = U 3 3) A B C = 4) A B C = A 8 2 4 1 7 B 6 5) A B C = 6) A B C = C 5 7) A B C = 8) A B C = 9) A B C = Problem 5. Draw a Venn diagram and use the given information to fill in the number of elements in each region. 1) n(a) = 15 2) n(a B C) = 5 3) n(a C) = 13 4) n(a B ) = 9 5) n(b C) = 8 6) n(a B C ) = 21 7) n(b C ) = 3 8) n(b C) = 32
Sets 2.4 Infinite and Finite Sets Expanded Introduction to Mathematical Reasoning Page 49 We will only discuss this section briefly. There will be no solving problems from this section, but there may be problems that test your understanding of the material. This section is very interesting and worth discussing. Definition 2.4.1: A set A is said to be finite if n(a) = x N of n(a) = 0. Definition 2.4.2: A set is said to be infinite if it can be placed into a 1-1 correspondence with a proper subset of itself. This asks us to pair elements together. That is one element in a set A is paired with one element from set B where B A. Notice that this definition eliminates the possibility that a finite set is infinite. This is because no finite set can be paired up 1-1 with a proper subset of its self. Ex: 2.4.1: Let A = {1, 2, 3}. Are there any proper subsets of A that have 1-1 correspondence with A? There are none. The proper subsets of A are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},. We know that there a total of 2 3 1 = 7 proper subsets of A, which we have. Notice that all of these sets have cardinality less than n(a). Hence it is impossible to have 1-1 correspondence. Ex: 2.4.2: Is the set N finite or infinite? We know that N = {1, 2, 3, 4,...}. Consider the set of even natural numbers B = {2, 4, 6, 8...}. Notice that the pairs can be given by the following N={ 1, 2, 3, 4,... } B={ 2, 4, 6, 8,... } This is a weird property. Even though every element of B is inside of N, we can still make a 1-1 correspondence because of the infinite amount of elements. Theorem 2.4.1: 2.4.1 Countable and Uncountable If you can make a 1-1 correspondence with N the set is a special infinite, called countable, or countably infinite. This is denoted by ℵ 0 (aleph-naught). Ex: 2.4.3: Give some examples of infinite sets using the common math sets and determine if they are countable or not. Some examples are Z, N, Q, Q, and R. We have already shown that N is countable. What about the rest? The integers is pretty easy to show, since it is exactly like the natural numbers. N={ 1, 2, 3, 4, 5,... } Z={ 0, -1, 1, -2, 2,... } The tricky one is Q. What would happen if we made a 1-1 correspondence as such N={ 1, 2, 3, 4, 5,... } 1 { 1, 1 2, 1 3, 1 4, 1 5,... } Notice that we would never be able to change the numerator. This causes a problem, that can be easily be fixed with some creativity.
Sets Expanded Introduction to Mathematical Reasoning Page 50 We can solve this by write the set Q out and moving in a diagonal fashion. We do this as such, 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6...... This allows us to move out of the first row. Each red line represents one element of the set N and gives us the 1-1 correspondence. What about Q and R? We won t show it, but they are not countable. However they are infinite. If they are not countably infinite, what could they be? These sets are called uncountably infinite or uncountable.since it is not countable, the cardinality is not ℵ 0. It happens to be the case that there are different ℵ numbers that represent cardinality. This also means that there are different sized infinities. Do you think that Q and R are the same size? How about Q and R? What about the set of all subsets of R and R itself?
Expanded Introduction to Mathematical Reasoning Page 51 2.4.2 Homework 2.4 Problem 1. Can you provide a real life example of an infinite set? Justify your answer. Problem 2. True or False. Provide explanations. 1. If A B and B is an infinite set, then A is infinite. True False 2. Every subset of a finite set is also finite. True False 3. Since N is an infinite set and R is an infinite set, n(n) = n(r). True False 4. The intersection of infinite sets is infinite. True False 5. The union of infinite sets is infinite. True False 6. The complement of is infinite. True False
Counting Expanded Introduction to Mathematical Reasoning Page 52 12 The Nature of Counting 12.1 Permutations and Counting The goal of this section is to give us the tools to be able to handle the probability section. We will learn two different different ways to count. The first is called the fundamental counting principle. Theorem 12.1.1: 12.1.1 Fundamental Counting Principle The fundamental counting principle gives the number of ways to perform a multiple number of tasks. If task A can be performed in m different ways and B can be performed in n different ways, then performing task A then task B can be done in m n different ways. This principle holds for more than 2 tasks. Ex: 12.1.1: Let A be a group with 3 people, B be a group with 3 people, and C be a group with 2 people. I want to find all the possible ways to group people from group A, B, and C. What if I only want to know the number of ways to group people from A,B, and C? Label the people in A as 1,2, and 3, the people in B as a,b, and c, and the people in C as α and β. The number of ways to make groups is given below. A B C 1 a b c α β α β α β 2 3 a b c a b c α β α β α β α β α β α β Each red path designates a different way to group people from group A, B, and C. For example, if we follow the path on the top, we would have 1 from A, a from C, and α from C, forming the group (1,a,α). This is a relatively small sample which could be done through this tree graph. However it is much simpler to answer the question about the number of different groups by using theorem 12.1.1. By theorem 12.1.1, we have the total of different groups is given by multiplying the number of elements in each set. Hence, we get 3 3 = 18. If you were to count all the possible paths along the red lines above, you would come out to this same number. Notice, in each column of red lines, we are essentially adding a number of paths that is the same as the number of elements in that set, but we are doing this the same number of times as the number of elements in the previous set. The first column adds 3 paths, then
Counting Expanded Introduction to Mathematical Reasoning Page 53 the second column adds three paths three times, and lastly the 3 column adds 2 paths 3 times for each group of a, b and c. Ex: 12.1.2: Find the number of ways to form groups, where the groups are taken from A, B, then C. 1) n(a) = 4, n(b) = 6, and n(c) = 20 We will have 4 6 20 = 480 different groups. 3) A =, B = {1, 2, 3, 4, 5, 6}, and C = {dog, cat, pig} We will have 0 6 3 = 0 different groups. 2) n(a) = 10, n(b) = 10, and n(c) = 10 We will have 10 3 = 1000 different groups. 4) A = N, B = {1}, and C = {2} Since N is an infinite set, we will have an infinite number of different groups. This fundamental principle leads to the next topic called permutations. 12.1.2 Permutations Permutations in essence is the fundamental counting principle. We will first define permutations and its notation. Theorem 12.1.2: A permutation of r elements from a set A with n elements is the number of different arrangements of those r elements selected without repetition. Order matters as well. This says that group (1,2) is different than group (2,1). Permutations is denoted at np r. n! np r = where n! = n(n 1)(n 2)... (3)(2)(1) (n r)! 0! = 1 Think about the fundamental counting principle. It says to take the number of elements in the first group and multiply it to the number of elements in the second and so on. Well, in permutations consider taking the set A = {1, 2, 3, 4, 5}, and we want to select a group of 2 where we do not allow repetition. How many total groups of 2 are there? Lets use the fundamental counting principle to see how this works. The first selection is from A with 5 elements, but the second choice is from A minus the element we chose first. This means that in the second choice we can only choose from 4 elements. By theorem 12.1.1, we have 5 4 = 20 groups of 2. Notice however that we will get the same answer using 5P 2 = 5! (5 2)! = 5 4 3 2 = 5 4 = 20. 3 2 Essentially the top n! gives you all possible choices. Then by dividing by (n r)!, we remove all the repeated versions where order mattered. Ex: 12.1.3: In a math class, the professor says that he will only give out 3 A s which will be distributed at random to a class of 30 students. The first person picked gets an A+, the second gets an A, and the third gets an A-. Regardless of how you do in the class the only students who get A s will be these three students. How many different ways can he give out these A s? We first need to determine if replacement are happens. Can the professor pick an A then put it back into the running for the next person? Why or why not? He cannot do this because he said that he would give out only one A. Next determine if order matters. This one is a little clearer, because the first person picked gets as different
Counting Expanded Introduction to Mathematical Reasoning Page 54 grade than the other two. Because we satisfy these conditions, we use permutations. 30P 3 = 30! 30 29... 3 2 = = 30 29 28 = 24, 360. (30 3)! 27 26... 3 2 There are 24,360 ways for the professor to give out the A s. Lets think about another situation. What would happen if we wanted to pick a group of size zero? How many different ways can we do this? Ex: 12.1.4: We have a group of 5 dogs staying at a kennel. The owner of the kennel want to know how many ways there are to group 0 dogs, assuming that the order matters. What is the answer? One way to think of this is, to say we want to have a group of zero dogs and there is only one way to do this. That is get a group of zero dogs. Another way to show this is mathematically. We know that we cannot pick a dog twice and order matters, hence we use permutations. By theorem 12.1.2, 5P 0 = There is only one way to group 0 dogs. 5! (5 0)! = 5! 5! = 1. 12.1.3 Distinguishable Permutations What if we want to find the number of ways to permute the letters of Hawaii. Can you tell the difference between the arrangement waiiah and waiiah? Well one had waiiah while the other waiiah. Notice we cannot determine the difference between these two arrangements. We need a new theorem to find these. Theorem 12.1.3: The number of distinguishable permutations of n objects in which n 1 is the number of objects of one kind, n 2 is the number of objects of another kind,..., and n k is the number of objects of a further kind, is given by n! n 1! n 2!... n k!. Ex: 12.1.5: How many different permutations of the letters of the given words are there? 1) Hawaii Notice a and i are repeated two times. Thus n 1 = 2 and n 2 = 2 and we have 6! 2!2! = 6 5 4 3 2 = 180 4 There are 180 distinguishable permutations of the letters of Hawaii. 2) Maui No letters are repeated so the denominator is 1. 4! 1! = 4 3 2 = 24 This is the same as 4 P 4. There are 24 distinguishable permutations of the letters of Maui.
Counting Expanded Introduction to Mathematical Reasoning Page 55 3) Kauai The only repeat is the a, which is repeated 2 times. Hence 5! 2! = 5 4 3 2 = 120 2 There are 120 distinguishable permutations of the letters of Kauai. 4) Humuhumunukunukuāpua a This one has a lot of letters repeated letters. Humuhumunukunukuāpua a There are 2 h, 9 u, 2 m, 2 n, 2 k, 3 a, and 1 p. Hence the formula gives us 21! = 1, 466, 593, 128, 000 2!9!2!2!2!3! There are 1,466,593,128,000 distinguishable permutations of the letters of Humuhumunukunukuāpua a.
Expanded Introduction to Mathematical Reasoning Page 56 12.1.4 Homework 12.1 Problem 1. Create a tree diagram showing all the possible outcomes of tossing three fair coins. Then list the ways of getting each result. 1) More than three tails. 2) Fewer than 2 tails. 3) At least 2 tails. 4) No more than 2 tails. Problem 2. Evaluate the expression without a calculator. 1) 16! 14! 2) 5! (5 2)! 3) 8! 6!(8 6)! 4) 100! 99!(100 99)! Problem 3. Find the number of distinguishable arrangements of the letters of each word or phrase. 1) GOOGOL 2) BANANA
Expanded Introduction to Mathematical Reasoning Page 57 Problem 4. How many five digit numbers are there in our system of counting numbers. Problem 5. True or False. Provide explanation. 1. (5 2)! 3! 5! = 0 2. The amount of ways to arrange the letters Lallay is 6! 2!2!. 3. n! (n 1)! is always n. Problem 6. Evaluate each expression. 1) 14 P 9 2) 11 P 3 3) 12 P 9 4) 9 P 3
Expanded Introduction to Mathematical Reasoning Page 58 Problem 7. First, second, and third prizes are to be awarded to three different people. If there are ten eligible candidates, how many outcomes are possible. Problem 8. At a wedding reception, the bride, the groom, and four attendants will form a reception line. How many ways can they be arranged in each of the following cases? 1. Any order will do. 2. The bride and groom must be the last two in line. 3. The groom must be the last in line with the bride next to him.
Counting 12.2 Combinations Expanded Introduction to Mathematical Reasoning Page 59 The book states that the permutations are without repetitions. They also hint toward the idea that combinations is opposite or with repetition. This is not the case. Combinations also must be without repetition. Combinations is discussing the amount aways to group thing, given that the order does not matter. For example, making a team of James and Cathy is the same as making a team of Cathy and James. Theorem 12.2.1: 12.2.1 Combinations Vs Permutations A combination of r elements from a set A with n elements is the number of different arrangements of those r elements selected without repetition with the property that order does matter. This says that group (1,2) is the same as group (2,1). Combinations is denoted at n C r. ( ) n n! nc r = = with r n. r r!(n r)! Ex: 12.2.1: You are the president of the math club. The club has members Cathy, James, Rob, Mat, and Halley. We want to form a three person committee. How many possible ways can we do this? 1. Is there repeats? No, we would not have a three person committee if James was selected twice. 2. Does order matter? No order does not matter because a committee of James, Cathy, and Rob is the same as Cathy, James, and Rob or Rob Cathy, and James. These are the questions that need to be asked every time there is a counting problem. We will give a flow chart later. With this, we will use combination to get ( ) 5 5! 5C 3 = = 3 3!(5 3)! = 5! 3!2! = 5 4 3 2 = 5 2 = 10. 3 2 2 There are 10 possible three person committees. What if the question was stated a little differently. We want to chose a secretary, treasurer, and a vice president. How many ways can we do this? Notice that the answer to question 1 is still the same. However, the order of which we pick members will be important. This tells us to use permutations. Hence, there are 5P 3 = 5! (5 3)! = 5! 2! = 5 4 3 2 = 5 4 3 = 60 2 possible ways to make a three person committee consisting of a secretary, treasurer, and a vice president. Does this make sense? The permutations is larger than the combinations, since order matter, we should have more possibilities. Now our previous example of James, Cathy, and Rob versus Cathy, James, and Rob or Rob Cathy, and James are all different. In combinations these would be worth one, but in permutations this is worth 3. Some of the most commonly found examples involve cards. Ex: 12.2.2: How many 5 card hands can you make from a standard deck of cards. 1. Is there repeats? No, we cannot have 2 kings of hearts with a standard deck of
Counting cards. Expanded Introduction to Mathematical Reasoning Page 60 2. Does order matter? No, because a 5 card hand 2, 3, 4, 5, and 6 is the same as 6, 5, 4, 3, and 2. Hence we use combinations and get ( ) 52 52! 52C 5 = = 5 5!(52 5)! = 52! 52 51 50 49 48 = = 2, 598, 960. 3!47! 3 2 We have 2,598,960 possible 5 card hands. Ex: 12.2.3: How many ways can you get a full house of three tens and two queens? Notice that this is a problem that requires both the fundamental counting principle as well as combinations. First we think about the different events discussed. Let A=Get three tens and B=Get two queens. By theorem 12.1.1 we get n(a) n(b) = Ways to get three tens Ways to get 2 queens. Now each of these parts is a counting problem in its own right. But they are solved the same way. For the first ways to get three tens, 1. Is there repeats? No, similarly as example 12.2.2. 2. Does order matter? No, similarly as example 12.2.2. So we have ( ) 4 3 ( ) 4 = 2 4! 3!(4 3)! 4! 2!(4 2)! = 4 3 2 3 2 1 4 3 2 = 4 2 3 = 24 2 2 There are a total of 24 ways to get a full house of 3 tens and 2 queens. 12.2.2 Binomial Theorem I won t discuss this to much, but if you would like to read more about it, you can find information in your book p. 578-579 or http://www.purplemath.com/modules/binomial. htm. The binomial theorem can be used to count, but we will be using it in a different way. First we will state the theorem and then give an example of its application. Theorem 12.2.2: For any positive integer n, n Z, ( ) ( ) n n (a + b) n = a n b 0 + a n 1 b + 0 1 ( ) ( ) n n a n 2 b 2 +... + ab n 1 + 2 n 1 ( ) n a 0 b n n Ex: 12.2.4: What is 3 7?
Counting Expanded Introduction to Mathematical Reasoning Page 61 Notice 3 7 can be rewritten as (1 + 2) 7. This is of the form of the right hand side of theorem 12.2.2. Hence, ( ) ( ) ( ) ( ) ( ) 7 7 7 7 7 (1 + 2) 7 = 1 7 2 0 + 1 6 2 1 + 1 5 2 2 + 1 4 2 3 + 1 3 2 4 0 1 2 3 4 ( ) ( ) ( ) 7 7 7 + 1 2 2 5 + 1 1 2 6 + 1 0 2 7 5 6 7 = (1)(1 7 )(1) + (7)(1 6 )(2) + 21(1 5 )(4) + 35(1 4 )(8) + 35(1 3 )(16) + 21(1 2 )(32) + 7(1 1 )(64) + 1(1 0 )(128) = 2187. This can be used to expand really large powers quickly. In previous math classes you have expanded problems like (x + 2) 2. What if the problem was (x + 2) 7? Evaluating (x + 2)(x + 2)(x + 2)(x + 2)(x + 2)(x + 2)(x + 2) would be very tedious. Where as evaluating ( ) 7 (x+2) 7 = x 7 2 ( ) 7 0 + x 6 2 ( ) 7 1 + x 5 2 ( ) 7 2 + x 4 2 ( ) 7 3 + x 3 2 ( ) 7 4 + x 2 2 ( ) 7 5 + x 1 2 ( ) 7 6 + x 0 2 7 0 1 2 3 4 5 6 7 = (1)(x 7 )(1)+(7)(x 6 )(2)+21(x 5 )(4)+35(x 4 )(8)+35(x 3 )(16)+21(x 2 )(32)+7(x 1 )(64)+1(x 0 )(128) is much easier.
Expanded Introduction to Mathematical Reasoning Page 62 12.2.3 Counting Flow Chart Straight to Fundamental Counting Principle Are items selected with replacement? Yes Use Fundamental Counting Principle No Does Order Matter? Yes Are Some Items Identical? Yes Use n! n 1! n 2!... n k! No No Use Combination Rule Permutations n! r!(n r)! n! (n r)!
Expanded Introduction to Mathematical Reasoning Page 63 Problem 1. Evaluate each expression. 12.2.4 Homework 12.2 1) 14 C 9 2) 10 C 8 3) 11 C 7 4) 12 C 9 Problem 2. First, second, and third prizes are to be awarded to three different people. If there are ten eligible candidates, how many outcomes are possible. Problem 3. Your boss has decided to start a project of building parks in inner cities. She hopes to bring youth programs to these areas in hopes of creating a safe space for children to be. She needs a team of 5 people to run the project and has tasked you to create the team. There are 7 people in your section and you need to make the team from your colleagues. How many possible teams are there?
Expanded Introduction to Mathematical Reasoning Page 64 Problem 4. Tyler is to build six homes on a block in a new subdivision, using two different models: standard and deluxe. (All standard model homes are the same, and all deluxe models are the same.) 1) How many different choices does Tyler have in positioning the six houses if he decides to build three standard and three deluxe models. 2) If Tyler builds two deluxe and four standards, how many different positioning can he use. Problem 5. In how many ways could eight people be divided into two groups of three people and a group of two people? Problem 6. Use the Binomial Theorem to evaluate 2 5. Problem 7. Use the Binomial Theorem to evaluate (x + 1) 5.
Counting 12.3 Counting Without Counting Expanded Introduction to Mathematical Reasoning Page 65 In this section we will be doing a multitude of problems where it is unknown which method of counting we will use. You will need to use section 12.2.3 to figure out if you need to use the fundamental counting principle, combinations, or permutations. The most difficult part about the upcoming probability section is counting. If you can count well, then the probability section will be much easier. Ex: 12.3.1: 12.3.1 Random Counting Problems How many different Hawaii state license plates are there? 1. Is there repeats? Yes, so we use the fundamental counting principle. Notice that Hawaii has a license plate ordered by 3 letters then 3 numbers. There are 26 letters in the alphabet and 10 possible numbers. By theorem 12.1.1, we have 26 26 26 10 10 10 = 26 3 10 3 = 17, 567, 000. There are 17,567,000 different license plates. Ex: 12.3.2: How many different Hawaii license plates are there if repetition is not allowed? We will do this in two different ways. First we will do this using theorem 12.1.1. The first task is to pick a letter which has 26 choices, the second task is to pick a letter which has 25, and the third task is to pick a letter which has 24, assuming no repetition. Also the first number or fourth task has 10 choices, the second number of fifth task has 9 choices, and the third number or sixth task has 8 choices. By theorem 12.1.1, we have 26 25 24 10 9 8 = 11, 232, 000. We may also do this using permutations since order of the letters matters and order of the numbers matters. We know that it is permutations since 1. Are there repeats? No. We cannot draw the same number or letter twice. 2. Does order matter? Yes. This is true since license plate ABC123 is not the same as BAC123. We would do this as Number of ways to get 3 letters The number of ways to get three numbers 26P 3 10 P 3 = 26! (26 3)! 10! (10 3)! = 26! 23! 10! = 26 25 24 10 9 8 = 11, 232, 000 7! We conclude there are 11,232,000 possible license plates without repetition. Ex: 12.3.3: How many different ticket numbers are there in the Powerball? The instructions for the Powerball are as follows. They draw 5 white balls from 69 different numbers and 1 red ball from 26 different balls numbered 1-26. You need to guess the right white numbers regardless of the order and the the red number correctly to win the jackpot. 1. Are there repeats? There are no repeats because the balls are taken out of the
Counting Expanded Introduction to Mathematical Reasoning Page 66 box, and they are not put back in. 2. Does order matter? No order does not matter. This is stated in the way that the Powerball is played. Hence we use combinations to get, 69C 5 26 = 69! 69 68 67 66 65 26 26 26 = = 292, 201, 338. 5!(69 5)! 5 4 3 2 The total number of ways to get a Powerball ticket is 292,201,338. Ex: 12.3.4: Lets say that the committee that runs the Powerball has found that to many people are winning. What is an easy way for them to fix this problem using what we know about counting? Instead of using combinations, where order does not matter, we would use permutations. This would increase the number of possibilities. 69P 5 26 = 69! (69 5)! 69 68 67 66 65 26 = 35, 064, 160, 560. The new Powerball system has a total of 35,064,160,560 different tickets. Another way to do this is to increase the cost of the Powerball tickets. This changes the expected value of the game, something that we will learn in the probability section. All this means is that the amount of money expected to go to the people running the game will increase. Ex: 12.3.5: How many ways are there to arrange the numbers 1,2,1, and 4. 1. Are there repeats? In other words even though there are two ones, the first one cannot be repeated twice. Nor could we get four ones. 2. Order matters, and some items are identical, hence we have a permutations problem where we need cannot distinguish between the arrangements 1142 and 1142. This says to use theorem 12.1.3 4! 2! = 4 3 2 = 12 2 There are 12 different arrangements of the numbers 1,2,1, and 4. Ex: 12.3.6: In a race with 30 runners where 8 trophies will be given to the top 8 runners (the trophies are distinct: first place, second place, etc), how many ways can this be done? 1. Is there repeats? No, since two runners cannot get a 1st place trophie. 2. Order matters, because the trophies are for different places. We are working with permutations, and we have 30P 8 = 30! = 30 29 28 27 26 25 24 23 = 235, 989, 936, 000. (30 8)!
Counting Expanded Introduction to Mathematical Reasoning Page 67 There are 235,989,936,000 different ways for 8 runners to win trophies with 30 total runners. Ex: 12.3.7: How many arrangements of the letters are there of the word repetition. We have the letters e,t, and i repeated twice. Using theorem 12.1.3, we have 10! 2!2!2! = 10 9 8... 4 3 2 = 10 9 8... 3 = 453, 600 2 4 There is a total of 453,600 arrangements of the letters of the word repetition. Ex: 12.3.8: An election ballot asks voters to select three city commissioners from a group of six candidates. In how many ways can this be done? 1. Is there repeats? Same person cannot be the first and second commissioner. His/her name cannot be placed back into the hat. 2. Order does not matter. This is just a group of three commissioners with no distinction between them. We use combinations to get 6! 3!(6 3)! = 6 5 4 3 2 = 5 4 = 20. 3 2 3 2 There are 20 different three person commissioner teams. The following problems were found online at http://idomath.weebly.com/uploads/7/ 0/9/8/7098279/permutationscombinationshw.pdf
Expanded Introduction to Mathematical Reasoning Page 68 12.3.2 Homework 12.3 Problem 1. In a race in which six automobiles are entered and there are no ties, in how many ways can the first four finishers come in? Problem 2. The model of the car you are thinking of buying is available in nine different colors and three different styles (hatchback, sedan, or station wagon). In how many ways can you order the car? Problem 3. A book club offers a choice of 8 books from a list of 40. In how many ways can a member make a collection? Problem 4. A medical researcher needs 6 people to test the effectiveness of an experimental drug. If 13 people have volunteered for the test, in how many ways can 6 people be selected? Problem 5. From a club of 20 people, in how many ways can a group of three members be selected?
Expanded Introduction to Mathematical Reasoning Page 69 Problem 6. From the 30 pictures I have of my daughter s first birthday, my digital picture frame will only hold 3 at a time. How many different groups of 3 pictures can I put on the frame? Problem 7. A popular brand of pen is available in three colors (red, green or blue) and four tips (bold, medium, fine, or micro). How many different choices of pens do you have with this brand? Problem 8. A corporation has ten members on its board of directors. In how many ways can it elect a president, vice-president, secretary and treasurer? Problem 9. For a segment of a radio show, a disc jockey can play 7 songs. If there are 12 songs to select from, in how many ways can the program for this segment be arranged? Problem 10. How many different ways can a director select 4 actors from a group of 20 actors to attend a workshop on performing in rock musicals?
Probability Expanded Introduction to Mathematical Reasoning Page 70 13 The Nature of Probability 13.1 Introduction to Probability Probability, as stated earlier, is essentially counting. We will explain probability using things that we have already learned in the counting section (section 12) as well as the set theory section (section 2). We first will need some definitions. Definition 13.1.1: An experiment is an observation of any physical occurrence. Definition 13.1.2: The sample space is the set of all possible outcomes or events of an experiment. You can think of this as the universal set U. Remember U changes depending on what we are talking about (definition 2.1.3). Definition 13.1.3: An event or outcome of an experiment are the possible things that can happen. Events are denoted using capitol letters. Ex: 13.1.1: Find the sample space and some of the possible events. 1. Flipping a coin. The sample space is S={Getting a heads, Getting a tails}. Each of these elements are events and we denote it as H=getting a heads and T =getting a tails. 2. Rolling a die. The sample space is S={Roll a 1, Roll a 2, Roll a 3, Roll a 4, Roll a 5, Roll a 6,}. The events are A=Roll a 1, B=Roll a 2, C=Roll a 3, D=Roll a 4, E=Roll a 5, and F =Roll a 6. Now since events are subsets of S, they may include more than one element. For example, in example 13.1.1 part 2, another event could be G=Roll a 4,5, or a 6. 13.1.1 Theoretical Probability First, when we talk about probability it is likely that I may forget to state fair, standard, or unbiased when discussing this topic. You can assume that the items that we are using are fair or unbiased unless stated otherwise. Theoretical probability is well theoretical. Every one knows that the probability of getting a heads when flipping a coin is 1 2. Does this necessarily mean that if I flip two coins, that one must be a head and the other must be a tail? Of course not. Probability is the concept of likeliness, not what must happen. Definition 13.1.4: Let E be an event. The theoretical probability of E if the ratio of ways to get into event E divided by the number of outcomes. Assuming that all the events have the same likelihood of happening. Theorem 13.1.1: P (E) = Ways to get into event E The number of outcomes = n(e) n(s) Properties of probability: Let E be an event within the sample space S. That is E is a subset of S. Then the following properties hold. 1. 0 P (E) 1 The probability of an event is a number from 0 through 1. 2. P ( ) = 0 The probability of an impossible event is 0. 3. P (S) = 1 The probability of a certain event is 1.
Probability Note: The events and the sample space relate. The sample space is physically what you did in the event. For example, if E is roll a 6, then S is roll a dice. However if E is roll a 6 and get a heads in coin flip, then S is rolling a dice and flipping a coin. 4. P (E) + P (E ) = 1. Instructions Expanded Introduction to Mathematical Reasoning Page 71 1. Write down the event(s) (What you want) and S physically what you are doing. 2. Write explicitly the ways to get into event E, or use section 12 to determine the ways to get into event E. 3. Write explicitly the ways to get into event S, or use section 12 to determine the ways to get into event S. 4. Write the formula. P (E) = 5. Plug in and simplify. Ways to get into event E The number of outcomes = n(e) n(s). Ex: 13.1.2: Find the probability of the following events. 1. What is the probability of getting a 4 when rolling a fair 6 sided dice? (a) Let E =Roll a 4. (b) There is only one way to do this i.e. roll a 4. So n(e) = 1. (c) There are six outcomes. Roll a 1,2,3,4,5 or 6. So n(s) = 6 (d) P (E) = is 1 6. Ways to get into event E The number of outcomes = n(e) n(s) = 1 6. The probability of rolling a 4 2. What is the probability of getting 1 head and 1 tail when flipping two fair coins? (a) Let E =Flip 1 head and 1 tail. (b) There are two ways to do this i.e. Flip a TH or HT. So n(e) = 2. (c) There are 4 possible outcomes. By theorem 12.1.1, the number of outcomes is 2 2 = 4. We can also list them as HH, HT, TH, or TT. So n(s) = 4 Ways to get into event E (d) P (E) = The number of outcomes = n(e) n(s) = 2 4 = 1 2. The probability of flipping two coins and getting 1 head and 1 tail is 1 2. 3. What is the probability of getting a queen when drawing a random card from a standard 52 card deck? (a) Let E =Draw a queen. (b) There are four ways to do this i.e. Q Q Q Q Q Q Q Q n(e) = 4. (c) There are 52 possible outcomes. This is just by the number of cards. So n(s) = 52 (d) P (E) = Ways to get into event E The number of outcomes drawing a queen is 1 13. = n(e) n(s) = 4 52 = 1 13. The probability of
Probability Expanded Introduction to Mathematical Reasoning Page 72 4. What is the probability of getting a queen or a king when drawing a random card from a standard 52 card deck? (a) Let E =Draw a queen or a king. (b) Don t forget that or is inclusive. Meaning we can draw a king, queen, or both at the same time. The last one is impossible, we only care about drawing a king or a queen separately. There are 8 ways to do this i.e. Q K Q K Q K K Q Q K K Q K Q K Q n(e) = 8. (c) There are still 52 possible outcomes. So n(s) = 52 (d) P (E) = Ways to get into event E The number of outcomes drawing a queen or a king is 2 13. 5. What is the probability of drawing a red 4? (a) Let E =Draw a red four. = n(e) n(s) = 8 52 = 2 13. The probability of (b) This is asking us to draw a card that is both red and 4. There are 2 ways to do this i.e. 4 4 4 4 n(e) = 2. (c) There are still 52 possible outcomes. So n(s) = 52 (d) P (E) = Ways to get into event E The number of outcomes drawing a red four is 1 26. = n(e) n(s) = 2 52 = 1 26. The probability of The following example is a very famous example which represents theorem 13.1.1 part 4. After reading this example, think about how this would play out in a game of deal or no deal. What should you do? Ex: 13.1.3: Monty Hall Problem: Suppose you re on a game show, and you re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, Do you want to pick door No. 2? Is it