Sets Chapter 2 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 1 2 4 1 Sets: A Problem Solving Tool 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 2.1 1 2 4 2 Sets A set is a well-defined collection of objects, called elements or members of the set. 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 Let A = {1, 2, 3, 4}, which means that “A is the set containing the elements 1, 2, 3, and 4.” 1 2 4 A “4 is in A” or “4 is an element of the set A” 6 A “6 is not an element of A” 4 3 3. Let X = {a, b, x, y}. Fill in the blank with or to make each statement correct. 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 a. a___X b. x___X c. A___X 1 2 4 4 Sets of Numbers Commonly used in Mathematics N = {x | x is a natural number} = {1, 2, 3, …} W = {x | x is a whole number} = {0,1, 2, 3, …} I = {x | x is an integer} = {…, 2, 1, 0,1, 2, 3, …} 1 2 Q = {x | x is a rational number} = {x | x, is of the form 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 a/b where a, b are integers and b 0} Q = {x | x is a rational number} = {x | x has a terminating or repeating decimal pattern} 4 S = {x | x is a irrational number} = {x | x is not rational} R = {x | x is a real number} = {x | x is the union of the rational and irrational numbers} 5 Describing Sets 1. Verbal or written description 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 2. Roster Method or Listing 3. Set-builder notation The Empty Set 1 2 The symbol { } or Ø represents the empty, or null set. 4 NOTE: {Ø} is not the null set. It is the set containing the null set. 6 Describe the following sets using the roster method and set builder notation. 00 11a.0 010 1 01set 0 110 00 01 01 00 numbers 1 011 The of1 counting less than 8. b. The set of counting numbers between 3 and 8. c. The set of counting numbers between 6 and 7. d. The set of counting numbers greater than 3. a. {1, 2, 3, 4, 5, 6, 7} {x | x is a counting number less than 8} number between 3 and 8} number between 6 and 7} number greater th an 3} b. {4, 5, 6, 7} {x | x is a counting c. { } or φ {x | x is a counting d. {4, 5, 6,…} {x | x is a counting 1 2 4 7 Equality of Sets Two sets A and B are equal, A = B, if they have the 00 11 0 010 1 01 0 110 1 00 01not 01 00 1 011 same elements necessarily listed in the same order. State whether the sets A and B are equal. 1. A={2n + 1|n is a counting number} B={2n – 1|n is a counting number} 2. A = {1,1,2,2,3} , B = {2,1,3} 1. A ≠ B , not equal 2. A = B , equal 1 2 4 8 Subsets The set A is a subset of B, A ⊆ B, if every element of A is also an element of B. 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 Proper Subsets A set A is said to be a proper subset of B, A ⊂ B, if A ⊆ B but A B. 1 2 4 The Universal Set The universal set, U, is the set of all elements under discussion. 9 List all the subsets and indicate which are the proper subsets of the given set. 00 11a.0 010 0 110 U1=01{a, b}1 00 01 01 00 1 011 b. U = {1, 2, 3} 1 a. Ø, {a}, {b}, and {a, b}. The first three are proper subsets. b. Ø {1}, {2}, {3} {1, 2}, {1, 3}, {2, 3} {1, 2, 3} The first seven are proper subsets. Note: by definition Ø is a subset of any set. 2 4 10 Number of Subsets of a Set n A set of n elements has 2 subsets. 00 11 0 010 1 01 0 110 1 00 01 01 00 1 011 Refer to the problems on the previous slide: 22 = 4 and 23 = 8. 1 2 Number of Proper Subsets of a Set A set of n elements has 2n – 1 proper subsets. 4 Refer to the problems on the previous slide: 22 1 = 4 1 = 3 and 23 1 = 8 1 = 7. 11 Example: Given: A = { 12, 19, 26,…, 68} Find a. the number of subsets 00 11 b. 0 010 1 01 0 110 1 00 01 01 00 1 011 the number of proper subsets To solve this problem you need to determine how many elements are in the set. Notice the elements in this set are evenly spaced and form what is called an arithmetic sequence. The next term is found by adding a common difference of seven in this example. The total number of elements can be found without having to list all of the elements in this set using the formula An = A1 + (n – 1)d. 1 2 An is the last term in the sequence, A1 is the first term, d is the common difference and n is the number of terms in the sequence. Substituting the numbers into the equation and solving for n yields the number of terms in this set. 68 = 12 + (n – 1)(7) –12 –12 56 = (n – 1)(7) 7 7 8 = n–1 +1 +1 9 = n 4 a. Number of subsets = 29 = 512 b. Number of proper subsets = 29 – 1 = 512 – 1 = 511 END 12