6.1 Logic. Statements or Propositions. Negation. The negation of a statement, p, is not p and is denoted by p Truth table: p p

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1 6.1 Logic Logic is not only the foundation of mathematics, but also is important in numerous fields including law, medicine, and science. Although the study of logic originated in antiquity, it was rebuilt and formalized in the 19 th and early 20 th century. George Boole (Boolean algebra) introduced mathematical methods to logic in 1847 while Georg Cantor did theoretical work on sets and discovered that there are many different sizes of infinite sets. Statements or Propositions A proposition or statement is a declaration which is either true or false. Some examples: 2+2 = 5 is a statement because it is a false declaration. Orange juice contains vitamin C is a statement that is true. Open the door. This is not considered a statement since we cannot assign a true or false value to this sentence. It is a command, but not a statement or proposition. Negation The negation of a statement, p, is not p and is denoted by p Truth table: p p T F F T If p is true, then its negation is false. If p is false, then its negation is true. 1

2 Disjunction A disjunction is of the form p V q and is read p or q. Truth table for disjunction: p q p V q T T T T F T F T T F F F A disjunction is true in all cases except when both p and q are false. Conjunction A conjunction is only true when both p and q are true. Otherwise, a conjunction of two statements will be false: Truth table: p q p q T T T T F F F T F F F F Conditional statement To understand the logic behind the truth table for the conditional statement, consider the following statement. If you get an A in the class, I will give you five bucks. Let p = statement You get an A in the class Let q = statement I will give you five bucks. Now, if p is true (you got an A) and I give you the five bucks, the truth value of p q is true. The contract was satisfied and both parties fulfilled the agreement. Now, suppose p is true (you got the A) and q is false (you did not get the five bucks). You fulfilled your part of the bargain, but weren t rewarded with the five bucks. So p q is false since the contract was broken by the other party. Now, suppose p is false. You did not get an A but received five bucks anyway. (q is true) No contract was broken. There was no obligation to receive 5 bucks, so truth value of p q cannot be false, so it must be true. Finally, if both p and q are false, the contract was not broken. You did not receive the A and you did not receive the 5 bucks. So p q is true in this case. 2

3 Truth table for conditional p q p q T T T T F F F T T F F T Variations of the conditional Converse: The converse of p q is q p Contrapositive: The contrapositive of p q p q is Examples Let p = you receive 90% Let q = you receive an A in the course p q? If you receive 90%, then you will receive an A in the course. Converse: q p If you receive an A in the course, then you receive 90% Is the statement true? No. What about the student who receives a score greater than 90? That student receives an A but did not achieve a score of exactly 90%. 3

4 Example 2 State the contrapositive in an English sentence: Let p = you receive 90% Let q = you receive an A in the course p q? If you receive 90%, then you will receive an A in the course q p If you don t receive an A in the course, then you didn t receive 90%. The contrapositive is true not only for these particular statements but for all statements, p and q. Logical equivalent statements p q p q Show that is logically equivalent to We will construct the truth tables for both sides and determine that the truth values for each statement are identical. The next slide shows that both statements are logically equivalent. The red columns are identical indicating the final truth values of each statement. 4

5 6.2 Sets This section will discuss the symbolism and concepts of set theory Set properties and set notation Definition of set: A set is any collection of objects specified in such a way that we can determine whether or not an object is or is not in the collection. Example 1. Set A is the set of all the letters in the alphabet. Notation: A = { a, b, c, d, e, z) We use capital letters to represent sets. We list the elements of the set within braces. The three dots indicate that the pattern continues. We can determine that an object is or is not in the collection. For example. e A stands for e is an element of, or e belongs to set A. This statement is true. The statement 3 A is false, since the number 3 is not an element of set A. The statement 3 A is true. Null set Example. What are the real number solutions of the equation? 2 x + 1= 0 Answer: There are no real number solutions of this equation since no real number squared added to one can ever equal 0. We represent the solution as the null set { } or Ø 1

6 Set builder notation Sometimes it is convenient to represent sets using what is called set-builder notation. For example, instead of representing the set A, letters in the alphabet by the roster method, we can use set builder notation: { x x is letter of the English alphabet} means the same as { a, b, c, d, e, z} 2 Example two. { x l x =9 } = {3, -3}. This is read as the set of all such that the square of x equals 9. The solution set consists of the two numbers 3 and -3. Subsets A B means A is a subset of B. A is a subset of A if every element of A is also contained in B. For example, the set of integers denoted by { -3, -2, -1, 0, 1, 2, 3, } is a subset of the set of real numbers. Formal definition of subset: A B means if x A, the x B (null set )is a subset of every set. To verify this statement, let s use the definition of subset. if x, then x is an element of A. But since the null set contains no elements, the statement x is an element of the null set is false. Hence, we have a conditional statement in which the premise is false. We know that p q is true if p is false. Since p is false, we conclude that the conditional statement is true. That is if x belongs to the null set, then x belongs to set A is true, which implies that the null set must be a member of every set. Therefore, the null set is a subset of every set. Number of subsets List all the subsets of set A = {bird, cat, dog} For convenience, we will use the notation A = {b, c, d} to represent set A. Solution: is a subset of A. We also know that every set is a subset of itself so A = {b, c, d } is a subset of set A since every element of set A is contained within set A. How many two-element subsets are there? We have {b, c}, {b, d}, {c, d} How many one-element subsets? { b}, {c} and {d}. There is a total of 8 subsets of set A if you count all the listed subsets. 2

7 Set operations The union of two sets is the set of all elements formed by combining all the elements of set A and all the elements of set B into one set. The symbolism used is The Venn Diagram representing the union of A and B is the entire region shaded yellow. A B A B= { x x Aor x B} Example of Union The union of the rational numbers with the set of irrational numbers is the set of real numbers. Rational numbers are those numbers that can be expressed as fractions, while irrational numbers are numbers that cannot be represented exactly as fractions, such as 2 Rational numbers a/b, ¾, 2/3, 0.6 Irrational numbers such as square root of two, Pi, square root of 3 Real numbers: represented by shaded blue-green region Intersection of sets A and B The intersection of sets A and B is the set of elements that is common to both sets A and B. It is symbolized as A B = { x l x A and x B } Represented by Venn Diagrams: A B Intersection 3

8 Complement of a set To understand the complement of a set, we must first define the universal set. The set of all elements under consideration is called the universal set. For example, when discussing numbers, the universal set may consist of the set of real numbers. All other types of numbers (integers, rational numbers, irrational numbers ) are subsets of the universal set of real numbers. Complement of set A: The complement of a set A is defined as the set of elements that are contained in U, the universal set, but not contained in set A. The symbolism for the complement of set A follows: A = { x U x A} Venn Diagram for complement of set A Yellow region= all The complement of set A is elements in U that represented by the regions are neither in A or A that are colored blue and B. yellow. The complement of set A is the region outside of the white circle representing set A. Elements of set B that are not in A A A B 4

9 6.3 Basic Counting Principles In this section, we will see how set operations play an important role in counting techniques. Opening example To see how sets play a role in counting, consider the following example. In a certain class, there are 23 majors in Psychology, 16 majors in English and 7 students who are majoring in both Psychology and English. If there are 50 students in the class, how many students are majoring in neither of these subjects? B) How many students are majoring in Psychology alone? Solution: We introduce the following principle of counting that can be illustrated using a Venn-Diagram. N( A U B) = n(a) + n(b) n(a B) A B This statement says that the number of elements in the union of two sets A and B is the number of elements of A added to the number of elements of B minus the number of elements that are in both A and B. 1

10 Do you see how the numbers of each region are obtained from the given information in the problem? We start with the region represented by the intersection of Psych. And English majors (7). Then, because there must be 23 Psych. Majors, there must be 16 Psych majors remaining in the rest of the set. A similar argument will convince you that there are 9 students who are majoring in English alone. 7 students in this region Both Psych and English 9 students in this region N(P U E) = n(p)+n(e)-n(p E) = 32 Psychology majors English majors 16 students here A second problem A survey of 100 college faculty who exercise regularly found that 45 jog, 30 swim, 20 cycle, 6 jog and swim, 1 jogs and cycles, 5 swim and cycle, and 1 does all three. How many of the faculty members do not do any of these three activities? How many just jog? We will solve this problem using a three-circle Venn Diagram in the accompanying slides. We will start with the intersection of all three circles. This region represents the number of faculty who do all three activities (one). Then, we will proceed to determine the number of elements in each intersection of exactly two sets J =joggers 1 does all 3 S=swimmers C=Cyclists 2

11 Solution: Starting with the intersection of all three circles, we place a 1 in that region (1 does all three). Then we know that since 6 jog and swim so 5 faculty remain in the region representing those who just jog and swim. Five swim and cycle, so 4 faculty just swim and cycle but do not do all three. Since 1 faculty is in the intersection region of joggers and cyclists, and we already have one that does all three activities, there must be no faculty who just jog and cycle. Multiplication principle dressed. The tree diagram illustrates the 24 ways to get To illustrate this principle, let s start with an example. Suppose you have 4 pairs of trousers in your closet, 3 different shirts and 2 pairs of shoes. Assuming that you must wear trousers (we hope so!), a shirt and shoes, how many different ways can you get dressed? Let s assume the colors of your pants are black, grey, rust, olive. You have four choices here. The shirt colors are green, marine blue and dark blue. For each pair of pants chosen (4) you have (3) options for shirts. You have 12 = 4*3 options for wearing a pair of trousers and a shirt. Now, each of these twelve options, you have two pair of shoes to choose from (Black or brown). Thus, you have a total of 4*3*2 = 24 options to get dressed. Generalized multiplication principle Suppose that a task can be performed using two or more consecutive operations. If the first operation can be accomplished in m ways and the second operation can be done in n ways, the third operation in p ways and so on, then the complete task can be performed in m n p ways.. 3

12 More problems How many different ways can a team consisting of 28 players select a captain and an assistant captain? Solution: Operation 1: select the captain. If all team members are eligible to be a captain, there are 28 ways this can be done. Operation 2. Select the assistant captain. Assuming that a player cannot be both a captain and assistant captain, there are 27 ways this can be done, since there are 27 team members left who are eligible to be the assistant captain. Then, using the multiplication principle there are (28)(27) ways to select both a captain and an assistant captain. This number turns out to be 756. Final example A sportswriter is asked to rank 8 teams in the NBA from first to last. How many rankings are possible? Solution: We will use 8 slots that need to be filled. In the first slot, we will determine how many ways to choose the first place team, the second slot is the number of ways to choose the second place team and so on until we get to the 8 th place team. There are 8 choices that can be made for the first place team since all teams are eligible. That leaves 7 choices for the second place team. The third place team is determined from the 6 remaining choices and so on Total is the product of 8(7) 1 =

13 6.4 Permutations and combinations For more complicated problems, we will need to develop two important concepts: permutations and combinations. Both of these concepts involve what is called the factorial of a number. Definition of n factorial (!) n! = n(n-1)(n-2)(n-3) 1 For example, 5! = 5(4)(3)(2)(1)=120 0! = 1 by definition. How it is used in counting: Example. The simplest protein molecule in biology is called vasopressin and is composed of 8 amino acids that are chemically bound together in a particular order. The order in which these amino acids occur is of vital importance to the proper functioning of vasopressin. If these 8 amino acids were placed in a hat and drawn out randomly one by one, how many different arrangements of these 8 amino acids are possible? Solution: Let A,B,C,D,E,F,G,H symbolize the 8 amino acids. They must fill 8 slots:. There are 8 choices for the first position, leaving 7 choices for the second slot, 6 choices for the third slot and so on. The number of different orderings is 8(7)(6)(5)(4)(3)(2)(1)=8! =40,320. Example continued: Of the 40,320 possible orderings of the 8 amino acids, the human body can use just one. What is the probability that, by random chance alone with no outside interference, the correct order occurs. We will discuss probability in the next chapter, but here is the answer: Probability of correct order is 1, an extremely unlikely event. 40,320 For more complicated biological molecules, such as hemoglobin, with many more amino acids, the probability that the correct order occurs by random chance alone is extremely small (close to zero!) which raises questions in some scientists minds of just how such molecules came to be formed by random chance. Some have concluded that their creation was not due to random chance but by intelligent design which raises still more questions that cannot be completely answered. 1

14 Two problems illustrating combinations and permutations. Consider the following two problems: 1) Consider the set { p, e, n} How many two-letter words (including nonsense words) can be formed from the members of this set? We will list all possibilities: pe, pn, en, ep, np, ne, a total of 6. 2) Now consider the set consisting of three males: {Paul, Ed, Nick} For simplicity, we will denote the set { p, e, n} How many two-man crews can be selected from this set? 3) Answer: pe (Paul, Ed), pn (Paul, Nick) and en (Ed, Nick) and that is all! Difference between permutations and combinations The difference between the two problems is this: Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. However, in the first problem, the order of the arrangements mattered since pe and ep are two different words. In the second problem, the order did not matter since pe and ep represented the same two-man crew. So we counted this only once. The first example was concerned with counting the number of permutations of 3 objects taken 2 at a time. The second example was concerned with the number of combinations of 3 objects taken 2 at a time Permutations The notation P(n,r) represents the number of permutations (arrangements) of n objects taken r at a time when r is less than or equal to n. In a permutation, the order is important. In our example, we have P(3,2) which represents the number of permutations of 3 objects taken 2 at a time. In our case, P(3,2) = 6 = (3)(2) In general, P(n,r) = n(n-1)(n-2)(n-3) (n-r+1) 2

15 More examples Use the definition P(n,r) = n(n-1)(n-2)(n-3) (n-r+1) Find P(5,3) Here, n = 5 and r = 3 so we have P(5,3) = (5)(5-1)5-3+1) = 5(4)3 = 60. This means there are 60 arrangements of 5 items taken 3 at a time. Application: How many ways can 5 people sit on a park bench if the bench can only seat 3 people? Solution: Think of the bench as three slots. There are five people that can sit in the first slot, leaving four remaining people to sit in the second position and finally 3 people eligible for the third slot. Thus, there are 5(4)(3)=60 ways the people can sit. The answer could have been found using the permutations formula: P(5,3) = 60, since we are finding the number of ways of arranging 5 objects taken 3 at a time. P(n,n)= n(n-1)(n-2) 1 Find P(5,5), the number of arrangements of 5 objects taken 5 at a time. Answer: P(5,5) = 5(5-1) (5-5+1) = 5(4)(3)(2)(1)=120. Application: A bookshelf has space for exactly 5 books. How many different ways can 5 books be arranged on this bookshelf? Think of 5 slots, again. There are five choices for the first slot, 4 for the second and so on until there is only 1 choice for the final slot. The answer is 5(4)(3)(2)(1) which is the same as P(5,5) = 120. Combinations In the second problem, the number of 2 man crews that can be selected from {p,e,n} was found to be 6. This corresponds to the number of combinations of 3 objects taken 2 at a time or C(3,2). We will use a variation of the formula for permutations to derive a formula for combinations. Consider the six permutations of { p, e, n} which are grouped in three pairs of 2. Each pair corresponds to one combination of 2. pe pn en ep np ne, so if we want to find the number of combinations of 3 objects taken 2 at a time, we simply divide the number of permutations of 3 objects taken 2 at a time by 2 (or 2!) P(3,2) We have the following result: C(3,2) = 2! 3

16 Generalization General result: This formula gives the number of subsets of size r that can be taken from a set of n objects. The order of the items in each subset does not matter. Pnr (, ) nn ( 1)( n 2)...( n r+ 1) Cnr (, ) = = r! r( r 1)( r 2)...1 Examples Find C(8,5) Solution: C(8,5) = P(8,5) 8(7)(6)(5)(4) 8(7)(6) = = = 8(7) = 56 5! 5(4)(3)(2)(1) 3(2)(1) 2. Find C(8,8) Solution: C(8,8) = P(8,8) 8(7)(6)(5)(4)(3)(2)(1) = = 1 8! 8(7)(6)(5)(4)(3)(2)(1) Combinations or Permutations? 1. In how many ways can you choose 5 out of 10 friends to invite to a dinner party? Solution: Does the order of selection matter? If you choose friends in the order A,B,C,D,E or A,C,B,D,E the same set of 5 was chosen, so we conclude that the order of selection does not matter. We will use the formula for combinations since we are concerned with how many subsets of size 5 we can select from a set of 10. C(10,5) = P(10,5) 10(9)(8)(7)(6) 10(9)(8)(7) = = = 2(9)(2)(7) = 252 5! 5(4)(3)(2)(1) (5)(4) 4

17 Permutations or Combinations? How many ways can you arrange 10 books on a bookshelf that has space for only 5 books? Does order matter? The answer is yes since the arrangement ABCDE is a different arrangement of books than BACDE. We will use the formula for permutations. We need to determine the number of arrangements of 10 objects taken 5 at a time so we have P(10,5) = 10(9)(8)(7)(6)=30,240 Lottery problem A certain state lottery consists of selecting a set of 6 numbers randomly from a set of 49 numbers. To win the lottery, you must select the correct set of six numbers. How many possible lottery tickets are there? Solution. The order of the numbers is not important here as long as you have the correct set of six numbers. To determine the total number of lottery tickets, we will use the formula for combinations and find C(49, 6), the number of combinations of 49 items taken 6 at a time. Using our calculator, we find that C(49,6) = 13,983,816 5