{"id":160,"date":"2016-01-25T22:05:39","date_gmt":"2016-01-25T22:05:39","guid":{"rendered":"https:\/\/courses.candelalearning.com\/math4libarts\/?post_type=chapter&#038;p=160"},"modified":"2021-06-24T19:27:36","modified_gmt":"2021-06-24T19:27:36","slug":"module-3-overview","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-mathforliberalarts\/chapter\/module-3-overview\/","title":{"raw":"Set Theory Basics","rendered":"Set Theory Basics"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe memberships of sets, including the empty set, using proper notation, and decide whether given items are members and determine the cardinality of a given set<\/li>\r\n \t<li>Perform the operations of union, intersection, complement, and difference on sets using proper notation<\/li>\r\n \t<li>Describe the relations between sets regarding membership, equality, subset, and proper subset, using proper notation<\/li>\r\n \t<li>Be able to draw and interpret Venn diagrams of set relations and operations and use Venn diagrams to solve problems<\/li>\r\n \t<li>Recognize when set theory is applicable to real-life situations, solve real-life problems, and communicate real-life problems and solutions to others<\/li>\r\n<\/ul>\r\n<\/div>\r\nIt is natural for us to classify items into groups, or sets, and consider how those sets overlap with each other. We can use these sets understand relationships between groups, and to analyze survey data.\r\n<h2>Introduction to Set Theory<\/h2>\r\nAn art collector might own a collection of paintings, while a music lover might keep a collection of CDs. Any collection of items can form a <strong>set<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Set<\/h3>\r\nA <strong>set<\/strong> is a collection of distinct objects, called <strong>elements<\/strong> of the set\r\n\r\nA set can be defined by describing the contents, or by listing the elements of the set, enclosed in curly brackets.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSome examples of sets defined by describing the contents:\r\n<ol>\r\n \t<li>The set of all even numbers<\/li>\r\n \t<li>The set of all books written about travel to Chile<\/li>\r\n<\/ol>\r\nSome examples of sets defined by listing the elements of the set:\r\n<ol>\r\n \t<li>{1, 3, 9, 12}<\/li>\r\n \t<li>{red, orange, yellow, green, blue, indigo, purple}<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Notation<\/h3>\r\nCommonly, we will use a variable to represent a set, to make it easier to refer to that set later.\r\n\r\nThe symbol \u2208\u00a0means \u201cis an element of\u201d.\r\n\r\nA set that contains no elements, { }, is called the <strong>empty set<\/strong> and is notated \u2205\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet <em>A <\/em>= {1, 2, 3, 4}\r\n\r\nTo notate that 2 is element of the set, we\u2019d write 2 \u2208 <em>A<\/em>\r\n\r\n<\/div>\r\nA set simply specifies the contents; order is not important. The set represented by {1, 2, 3} is equivalent to the set {3, 1, 2}.\r\n<h3>Subsets<\/h3>\r\nSometimes a collection might not contain all the elements of a set. For example, Chris owns three Madonna albums. While Chris\u2019s collection is a set, we can also say it is a <strong>subset<\/strong> of the larger set of all Madonna albums.\r\n<div class=\"textbox\">\r\n<h3>Subset<\/h3>\r\nA <strong>subset<\/strong> of a set <em>A<\/em> is another set that contains only elements from the set <em>A<\/em>, but may not contain all the elements of <em>A<\/em>.\r\n\r\nIf <em>B<\/em> is a subset of <em>A<\/em>, we write <em>B<\/em>\u00a0\u2286\u00a0<em>A<\/em>\r\n\r\nA <strong>proper subset<\/strong> is a subset that is not identical to the original set\u2014it excludes at least one element of the original set.\r\n\r\nIf <em>B<\/em> is a proper subset of <em>A<\/em>, we write <em>B<\/em> \u2282 <em>A<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConsider these three sets:\r\n\r\n<em>A<\/em> = the set of all even numbers\r\n<em>B<\/em> = {2, 4, 6}\r\n<em>C<\/em> = {2, 3, 4, 6}\r\n\r\nHere <em>B<\/em> \u2282 <em>A<\/em> since every element of <em>B<\/em> is also an even number, so is an element of <em>A<\/em>.\r\n\r\nMore formally, we could say <em>B<\/em> \u2282 <em>A<\/em> since if <em>x <\/em>\u2208\u00a0<em>B<\/em>, then <em>x <\/em>\u2208 <em>A<\/em>.\r\n\r\nIt is also true that <em>B<\/em> \u2282 <em>C<\/em>.\r\n\r\n<em>C<\/em> is not a subset of <em>A<\/em>, since C contains an element, 3, that is not contained in <em>A<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125855&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/5xthPHH4i_A?list=PL7138FAEC01D6F3F3\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose a set contains the plays \u201cMuch Ado About Nothing,\u201d \u201cMacBeth,\u201d and \u201cA Midsummer\u2019s Night Dream.\u201d What is a larger set this might be a subset of?\r\n[reveal-answer q=\"42047\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"42047\"]There are many possible answers here. One would be the set of plays by Shakespeare. This is also a subset of the set of all plays ever written. It is also a subset of all British literature.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConsider the\u00a0set\u00a0[latex]A = \\{1, 3, 5\\} [\/latex]. Which of the following sets is [latex]A [\/latex] a subset of?\r\n[latex]X = \\{1, 3, 7, 5\\} [\/latex]\r\n[latex]Y = \\{1, 3 \\} [\/latex]\r\n[latex]Z = \\{1, m, n, 3, 5\\}[\/latex]\r\n[reveal-answer q=\"3546\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3546\"] [latex] X [\/latex] and [latex] Y [\/latex] [\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nGiven the\u00a0set: <em>A<\/em> = {<em>a<\/em>, <em>b<\/em>, <em>c<\/em>, <em>d<\/em>}. List all of the subsets of <em>A\r\n<\/em>[reveal-answer q=\"706217\"]Show Solution[\/reveal-answer]<em>\r\n<\/em>[hidden-answer a=\"706217\"]{} (or \u00d8), {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d},\u00a0{b,c,d}, {a,b,c,d}\r\n\r\nYou can see that there are 16 subsets, 15 of which are proper subsets.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nListing the sets is fine if you have only a few elements. However, if we were to list all of the subsets of a set containing many elements, it would be quite tedious. Instead, in the next example we will consider each element of the set separately.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn the previous example, there are four elements. For the first element, <em>a<\/em>, either it\u2019s in the set or it\u2019s not. Thus there are 2 choices for that first element. Similarly, there are two choices for <em>b<\/em>\u2014either it\u2019s in the set or it\u2019s not. Using just those two elements,\u00a0list all the possible subsets of the set {a,b}\r\n[reveal-answer q=\"857946\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"857946\"]\r\n\r\n{}\u2014both elements are not in the set\r\n{<em>a<\/em>}\u2014<em>a<\/em> is in;\u00a0<em>b<\/em> is not in the set\r\n{<em>b<\/em>}\u2014<em>a<\/em> is not in the set;\u00a0<em>b<\/em> is in\r\n{<em>a<\/em>,<em>b<\/em>}\u2014<em>a<\/em> is in; <em>b<\/em> is in\r\n\r\nTwo choices for <em>a<\/em>\u00a0times the two for <em>b<\/em>\u00a0gives us [latex]2^{2}=4[\/latex] subsets.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow let\u2019s include <em>c,\u00a0<\/em>just for fun. List all the possible subsets of the new set {a,b,c}.\r\nAgain, either <em>c<\/em> is included or it isn\u2019t, which gives us two choices. The outcomes are {}, {<em>a<\/em>}, {<em>b<\/em>}, {<em>c<\/em>}, {<em>a<\/em>,<em>b<\/em>}, {<em>a<\/em>,<em>c<\/em>}, {<em>b<\/em>,<em>c<\/em>}, {<em>a<\/em>,<em>b<\/em>,<em>c<\/em>}. Note that there are [latex]2^{3}=8[\/latex] subsets.\r\n\r\nIf you include\u00a0four elements, there would be [latex]2^{4}=16[\/latex] subsets. 15 of those subsets are proper, 1 subset, namely {<em>a<\/em>,<em>b<\/em>,<em>c<\/em>,<em>d<\/em>}, is not.\r\n\r\nIn general, if you have <em>n<\/em> elements in your set, then there are [latex]2^{n}[\/latex] subsets and [latex]2^{n}\u22121[\/latex] proper subsets.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]132343[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Union, Intersection, and Complement<\/h2>\r\nCommonly, sets interact. For example, you and a new roommate decide to have a house party, and you both invite your circle of friends. At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets.\r\n<div class=\"textbox\">\r\n<h3>Union, Intersection, and Complement<\/h3>\r\nThe <strong>union<\/strong> of two sets contains all the elements contained in either set (or both sets).\u00a0The union is notated <em>A <\/em>\u22c3<em> B.\u00a0<\/em>More formally, <em>x <\/em>\u220a <em>A <\/em>\u22c3 <em>B<\/em> if <em>x <\/em>\u2208 <em>A<\/em> or <em>x <\/em>\u2208 <em>B<\/em> (or both)\r\n\r\nThe <strong>intersection <\/strong>of two sets contains only the elements that are in both sets.\u00a0The intersection is notated <em>A <\/em>\u22c2<em> B.\u00a0<\/em>More formally, <em>x <\/em>\u2208 <em>A <\/em>\u22c2 <em>B<\/em> if <em>x <\/em>\u2208 <em>A<\/em> and <em>x <\/em>\u2208 <em>B.<\/em>\r\n\r\nThe <strong>complement<\/strong> of a set <em>A<\/em> contains everything that is <em>not<\/em> in the set <em>A<\/em>.\u00a0The complement is notated <em>A\u2019<\/em>, or <em>A<sup>c<\/sup>, or sometimes ~<em>A<\/em>.<\/em>\r\n\r\nA <strong>universal set<\/strong> is a set that contains all the elements we are interested in. This would have to be defined by the context.\r\n\r\nA complement is relative to the universal set, so\u00a0<em>A<sup>c<\/sup><\/em><em>\u00a0<\/em>contains all the elements in the universal set that are not in <em>A<\/em>.\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<ol>\r\n \t<li>If we were discussing searching for books, the universal set might be all the books in the library.<\/li>\r\n \t<li>If we were grouping your Facebook friends, the universal set would be all your Facebook friends.<\/li>\r\n \t<li>If you were working with sets of numbers, the universal set might be all whole numbers, all integers, or all real numbers<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose the universal set is <em>U<\/em> = all whole numbers from 1 to 9. If <em>A<\/em> = {1, 2, 4}, then\u00a0<em>A<sup>c<\/sup> <\/em>= {3, 5, 6, 7, 8, 9}.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110368&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConsider the sets:\r\n\r\n<em>A<\/em> = {red, green, blue}\r\n<em>B<\/em> = {red, yellow, orange}\r\n<em>C<\/em> = {red, orange, yellow, green, blue, purple}\r\n\r\nFind the following:\r\n<ol>\r\n \t<li>Find <em>A <\/em>\u22c3<em> B<\/em><\/li>\r\n \t<li>Find <em>A <\/em>\u22c2<em> B<\/em><\/li>\r\n \t<li>Find <em>A<sup>c<\/sup><\/em>\u22c2<em> C<\/em><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"691926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"691926\"]\r\n<ol>\r\n \t<li>The union contains all the elements in either set: <em>A <\/em>\u22c3<em> B<\/em> = {red, green, blue, yellow, orange}\u00a0Notice we only list red once.<\/li>\r\n \t<li>The intersection contains all the elements in both sets: <em>A <\/em>\u22c2<em> B<\/em> = {red}<\/li>\r\n \t<li>Here we\u2019re looking for all the elements that are <em>not<\/em> in set <em>A<\/em> and are also in <em>C<\/em>.\u00a0<em>A<sup>c<\/sup> <\/em>\u22c2<em> C<\/em> = {orange, yellow, purple}<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125865&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"450\"><\/iframe>\r\n\r\n<\/div>\r\nNotice that in the example above, it would be hard to just ask for <em>A<sup>c<\/sup><\/em>, since everything from the color fuchsia to puppies and peanut butter are included in the complement of the set. For this reason, complements are usually only used with intersections, or when we have a universal set in place.\r\n\r\nAs we saw earlier with the expression\u00a0<em>A<sup>c<\/sup><\/em><em>\u00a0<\/em>\u22c2<em> C<\/em>, set operations can be grouped together. Grouping symbols can be used like they are with arithmetic \u2013 to force an order of operations.\r\n\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose <em>H<\/em> = {cat, dog, rabbit, mouse}, <em>F<\/em> = {dog, cow, duck, pig, rabbit}, and\u00a0<em>W<\/em> = {duck, rabbit, deer, frog, mouse}\r\n<ol>\r\n \t<li>Find (<em>H <\/em>\u22c2<em> F<\/em>) \u22c3<em> W<\/em><\/li>\r\n \t<li>Find <em>H <\/em>\u22c2 (<em>F<\/em> \u22c3<em> W<\/em>)<\/li>\r\n \t<li>Find (<em>H <\/em>\u22c2<em> F<\/em>)<em><sup>c<\/sup><\/em> \u22c2<em> W<\/em><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"444204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444204\"]\r\n<ol>\r\n \t<li>We start with the intersection: <em>H <\/em>\u22c2<em> F<\/em> = {dog, rabbit}.\u00a0Now we union that result with <em>W<\/em>: (<em>H <\/em>\u22c2<em> F<\/em>) \u22c3<em> W<\/em> = {dog, duck, rabbit, deer, frog, mouse}<\/li>\r\n \t<li>We start with the union: <em>F<\/em> \u22c3<em> W<\/em> = {dog, cow, rabbit, duck, pig, deer, frog, mouse}.\u00a0Now we intersect that result with <em>H<\/em>: <em>H <\/em>\u22c2 (<em>F<\/em> \u22c3<em> W<\/em>) = {dog, rabbit, mouse}<\/li>\r\n \t<li>We start with the intersection: <em>H <\/em>\u22c2<em> F<\/em> = {dog, rabbit}.\u00a0Now we want to find the elements of <em>W<\/em> that are <em>not<\/em> in <em>H <\/em>\u22c2<em> F.\u00a0<\/em>(<em>H <\/em>\u22c2<em> F)<sup>c<\/sup><\/em>\u00a0\u22c2<em> W<\/em> = {duck, deer, frog, mouse}<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Venn Diagrams<\/h3>\r\nTo visualize the interaction of sets, John Venn in 1880 thought to use overlapping circles, building on a similar idea used by Leonhard Euler in the 18th century. These illustrations now called <strong>Venn Diagrams<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Venn Diagram<\/h3>\r\nA Venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Overlapping areas indicate elements common to both sets.\r\n\r\nBasic Venn diagrams can illustrate the interaction of two or three sets.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCreate Venn diagrams to illustrate <em>A <\/em>\u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>, and <em>Ac <\/em>\u22c2<em> B<\/em>\r\n\r\n<em>A <\/em>\u22c3<em> B<\/em> contains all elements in <em>either<\/em> set.\r\n[reveal-answer q=\"252649\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252649\"]\r\n\r\n<img class=\"alignnone size-full wp-image-449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220329\/ab1.png\" alt=\"ab1\" width=\"179\" height=\"125\" \/>\r\n\r\n<em>A <\/em>\u22c2<em> B<\/em> contains only those elements in both sets \u2013 in the overlap of the circles.\r\n\r\n<img class=\"alignnone size-full wp-image-450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220349\/ab2.png\" alt=\"ab2\" width=\"180\" height=\"125\" \/>\r\n\r\n<em>Ac <\/em>will contain all elements <em>not<\/em> in the set A. <em>A<sup>c <\/sup><\/em>\u22c2<em> B<\/em> will contain the elements in set <em>B<\/em> that are not in set <em>A<\/em>.\r\n\r\n<img class=\"alignnone size-full wp-image-451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220403\/ab3.png\" alt=\"ab3\" width=\"180\" height=\"125\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse a Venn diagram to illustrate (<em>H <\/em>\u22c2<em> F<\/em>)<sup><em>c<\/em><\/sup> \u22c2<em> W\r\n<\/em>[reveal-answer q=\"311976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"311976\"]\r\n\r\nWe\u2019ll start by identifying everything in the set <em>H <\/em>\u22c2<em> F<\/em>\r\n\r\n<img class=\"alignnone size-full wp-image-452\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220617\/hfw1.png\" alt=\"hfw1\" width=\"180\" height=\"152\" \/>\r\n\r\nNow, (<em>H <\/em>\u22c2<em> F<\/em>)<em>c<\/em> \u22c2<em> W<\/em> will contain everything <em>not<\/em> in the set identified above that is also in set <em>W<\/em>.\r\n\r\n<img class=\"alignnone size-full wp-image-453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220632\/hfw2.png\" alt=\"hfw2\" width=\"180\" height=\"156\" \/>\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/CPeeOUldZ6M\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCreate an expression to represent the outlined part of the Venn diagram shown.\r\n\r\n<img class=\"alignnone size-full wp-image-454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220943\/hfw3.png\" alt=\"hfw3\" width=\"180\" height=\"157\" \/>\r\n[reveal-answer q=\"828282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"828282\"]\r\n\r\nThe elements in the outlined set <em>are<\/em> in sets <em>H<\/em> and <em>F<\/em>, but are not in set <em>W<\/em>. So we could represent this set as <em>H <\/em>\u22c2<em> F<\/em> \u22c2<em> W<sup>c<\/sup><\/em>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nCreate an expression to represent the outlined portion of the Venn diagram shown.\r\n\r\n<img class=\"alignnone size-full wp-image-455\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12221621\/abc1.png\" alt=\"abc1\" width=\"180\" height=\"157\" \/>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6699&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"200\"><\/iframe>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/VfC8rhLdYYg\r\n<h2>Cardinality<\/h2>\r\nOften times we are interested in the number of items in a set or subset. This is called the cardinality of the set.\r\n<div class=\"textbox\">\r\n<h3>Cardinality<\/h3>\r\nThe number of elements in a set is the cardinality of that set.\r\n\r\nThe cardinality of the set <em>A<\/em> is often notated as |<em>A<\/em>| or n(<em>A<\/em>)\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nLet <em>A<\/em> = {1, 2, 3, 4, 5, 6} and <em>B<\/em> = {2, 4, 6, 8}.\r\n\r\nWhat is the cardinality of <em>B<\/em>? <em>A<\/em> \u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>?\r\n[reveal-answer q=\"100844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100844\"]\r\n\r\nThe cardinality of <em>B<\/em> is 4, since there are 4 elements in the set.\r\n\r\nThe cardinality of <em>A<\/em> \u22c3<em> B<\/em> is 7, since <em>A<\/em> \u22c3<em> B<\/em> = {1, 2, 3, 4, 5, 6, 8}, which contains 7 elements.\r\n\r\nThe cardinality of <em>A <\/em>\u22c2<em> B<\/em> is 3, since <em>A <\/em>\u22c2<em> B<\/em> = {2, 4, 6}, which contains 3 elements.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=109846&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nWhat is the cardinality of <em>P<\/em> = the set of English names for the months of the year?\r\n[reveal-answer q=\"6805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"6805\"]\r\n\r\nThe cardinality of this set is 12, since there are 12 months in the year.\r\n\r\nSometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nA survey asks 200 people \u201cWhat beverage do you drink in the morning\u201d, and offers choices:\r\n<ul>\r\n \t<li>Tea only<\/li>\r\n \t<li>Coffee only<\/li>\r\n \t<li>Both coffee and tea<\/li>\r\n<\/ul>\r\nSuppose 20 report tea only, 80 report coffee only, 40 report both.\u00a0\u00a0 How many people drink tea in the morning? How many people drink neither tea or coffee?\r\n[reveal-answer q=\"432276\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"432276\"]\r\n\r\nThis question can most easily be answered by creating a Venn diagram. We can see that we can find the people who drink tea by adding those who drink only tea to those who drink both: 60 people.\r\n\r\nWe can also see that those who drink neither are those not contained in the any of the three other groupings, so we can count those by subtracting from the cardinality of the universal set, 200.\r\n\r\n200 \u2013 20 \u2013 80 \u2013 40 = 60 people who drink neither.\r\n\r\n<img class=\"alignnone size-full wp-image-458\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222016\/coffeetea.png\" alt=\"coffeetea\" width=\"180\" height=\"124\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125872&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"200\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA survey asks:\u00a0\u00a0 Which online services have you used in the last month:\r\n<ul>\r\n \t<li>Twitter<\/li>\r\n \t<li>Facebook<\/li>\r\n \t<li>Have used both<\/li>\r\n<\/ul>\r\nThe results show 40% of those surveyed have used Twitter, 70% have used Facebook, and 20% have used both. How many people have used neither Twitter or Facebook?\r\n\r\n[reveal-answer q=\"778879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"778879\"]\r\n\r\nLet <em>T<\/em> be the set of all people who have used Twitter, and <em>F<\/em> be the set of all people who have used Facebook. Notice that while the cardinality of <em>F<\/em> is 70% and the cardinality of <em>T<\/em> is 40%, the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> is not simply 70% + 40%, since that would count those who use both services twice. To find the cardinality of <em>F<\/em> \u22c3 <em>T<\/em>, we can add the cardinality of <em>F<\/em> and the cardinality of <em>T<\/em>, then subtract those in intersection that we\u2019ve counted twice. In symbols,\r\n\r\nn(<em>F<\/em> \u22c3 <em>T<\/em>) = n(<em>F<\/em>) + n(<em>T<\/em>) \u2013 n(<em>F<\/em> \u22c2 <em>T<\/em>)\r\n\r\nn(<em>F<\/em> \u22c3 <em>T<\/em>) = 70% + 40% \u2013 20% = 90%\r\n\r\nNow, to find how many people have not used either service, we\u2019re looking for the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> .\r\n\r\nSince the universal set contains 100% of people and the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> = 90%, the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> must be the other 10%.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe previous example illustrated two important properties called cardinality properties:\r\n<div class=\"textbox\">\r\n<h3>Cardinality properties<\/h3>\r\n<ol>\r\n \t<li>n(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)<\/li>\r\n \t<li>n(<em>Ac<\/em>) = n(<em>U<\/em>) \u2013 n(<em>A<\/em>)<\/li>\r\n<\/ol>\r\n<\/div>\r\nNotice that the first property can also be written in an equivalent form by solving for the cardinality of the intersection:\r\n<p style=\"text-align: center;\">n(<em>A<\/em> \u22c2 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c3 <em>B<\/em>)<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFifty students were surveyed, and asked if they were taking a social science (SS), humanities (HM) or a natural science (NS) course the next quarter.\r\n\r\n21 were taking a SS course\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 26 were taking a HM course\r\n\r\n19 were taking a NS course\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9 were taking SS and HM\r\n\r\n7 were taking SS and NS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 were taking HM and NS\r\n\r\n3 were taking all three\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7 were taking none\r\n\r\nHow many students are only taking a SS course?\r\n[reveal-answer q=\"88483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88483\"]\r\n\r\nIt might help to look at a Venn diagram.\r\n\r\n<img class=\"alignnone size-full wp-image-459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222214\/sshmns.png\" alt=\"sshmns\" width=\"180\" height=\"152\" \/>\r\n\r\nFrom the given data, we know that there are\r\n\r\n3 students in region <em>e<\/em> and\r\n\r\n7 students in region <em>h<\/em>.\r\n\r\nSince 7 students were taking a SS and NS course, we know that n(<em>d<\/em>) + n(<em>e<\/em>) = 7. Since we know there are 3 students in region 3, there must be\r\n\r\n7 \u2013 3 = 4 students in region <em>d<\/em>.\r\n\r\nSimilarly, since there are 10 students taking HM and NS, which includes regions <em>e<\/em> and <em>f<\/em>, there must be\r\n\r\n10 \u2013 3 = 7 students in region <em>f<\/em>.\r\n\r\nSince 9 students were taking SS and HM, there must be 9 \u2013 3 = 6 students in region <em>b<\/em>.\r\n\r\nNow, we know that 21 students were taking a SS course. This includes students from regions <em>a, b, d, <\/em>and <em>e<\/em>. Since we know the number of students in all but region <em>a<\/em>, we can determine that 21 \u2013 6 \u2013 4 \u2013 3 = 8 students are in region <em>a<\/em>.\r\n\r\n8 students are taking only a SS course.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nOne hundred fifty people were surveyed and asked if they believed in UFOs, ghosts, and Bigfoot.\r\n\r\n43 believed in UFOs\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 44 believed in ghosts\r\n\r\n25 believed in Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 believed in UFOs and ghosts\r\n\r\n8 believed in ghosts and Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5 believed in UFOs and Bigfoot\r\n\r\n2 believed in all three\r\n\r\nHow many people surveyed believed in at least one of these things?\r\n[reveal-answer q=\"252699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252699\"]\r\n\r\n1. There are several answers: The set of all odd numbers less than 10. The set of all odd numbers. The set of all integers. The set of all real numbers.\r\n\r\n2. <em>A <\/em>\u22c3<em> C<\/em> = {red, orange, yellow, green, blue purple}\r\n\r\n<em>Bc <\/em>\u22c2<em> A<\/em> = {green, blue}\r\n\r\n3. <em>A <\/em>\u22c3<em> B <\/em>\u22c2<em> Cc<\/em>\r\n\r\n4. Starting with the intersection of all three circles, we work our way out. Since 10 people believe in UFOs and Ghosts, and 2 believe in all three, that leaves 8 that believe in only UFOs and Ghosts. We work our way out, filling in all the regions. Once we have, we can add up all those regions, getting 91 people in the union of all three sets. This leaves 150 \u2013 91 = 59 who believe in none.\r\n\r\n<img class=\"alignnone size-full wp-image-460\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222338\/ufoghostsbigfoot.png\" alt=\"ufoghostsbigfoot\" width=\"183\" height=\"152\" \/>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom20\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125868&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/wErcETeKvrU","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe memberships of sets, including the empty set, using proper notation, and decide whether given items are members and determine the cardinality of a given set<\/li>\n<li>Perform the operations of union, intersection, complement, and difference on sets using proper notation<\/li>\n<li>Describe the relations between sets regarding membership, equality, subset, and proper subset, using proper notation<\/li>\n<li>Be able to draw and interpret Venn diagrams of set relations and operations and use Venn diagrams to solve problems<\/li>\n<li>Recognize when set theory is applicable to real-life situations, solve real-life problems, and communicate real-life problems and solutions to others<\/li>\n<\/ul>\n<\/div>\n<p>It is natural for us to classify items into groups, or sets, and consider how those sets overlap with each other. We can use these sets understand relationships between groups, and to analyze survey data.<\/p>\n<h2>Introduction to Set Theory<\/h2>\n<p>An art collector might own a collection of paintings, while a music lover might keep a collection of CDs. Any collection of items can form a <strong>set<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Set<\/h3>\n<p>A <strong>set<\/strong> is a collection of distinct objects, called <strong>elements<\/strong> of the set<\/p>\n<p>A set can be defined by describing the contents, or by listing the elements of the set, enclosed in curly brackets.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Some examples of sets defined by describing the contents:<\/p>\n<ol>\n<li>The set of all even numbers<\/li>\n<li>The set of all books written about travel to Chile<\/li>\n<\/ol>\n<p>Some examples of sets defined by listing the elements of the set:<\/p>\n<ol>\n<li>{1, 3, 9, 12}<\/li>\n<li>{red, orange, yellow, green, blue, indigo, purple}<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox\">\n<h3>Notation<\/h3>\n<p>Commonly, we will use a variable to represent a set, to make it easier to refer to that set later.<\/p>\n<p>The symbol \u2208\u00a0means \u201cis an element of\u201d.<\/p>\n<p>A set that contains no elements, { }, is called the <strong>empty set<\/strong> and is notated \u2205<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let <em>A <\/em>= {1, 2, 3, 4}<\/p>\n<p>To notate that 2 is element of the set, we\u2019d write 2 \u2208 <em>A<\/em><\/p>\n<\/div>\n<p>A set simply specifies the contents; order is not important. The set represented by {1, 2, 3} is equivalent to the set {3, 1, 2}.<\/p>\n<h3>Subsets<\/h3>\n<p>Sometimes a collection might not contain all the elements of a set. For example, Chris owns three Madonna albums. While Chris\u2019s collection is a set, we can also say it is a <strong>subset<\/strong> of the larger set of all Madonna albums.<\/p>\n<div class=\"textbox\">\n<h3>Subset<\/h3>\n<p>A <strong>subset<\/strong> of a set <em>A<\/em> is another set that contains only elements from the set <em>A<\/em>, but may not contain all the elements of <em>A<\/em>.<\/p>\n<p>If <em>B<\/em> is a subset of <em>A<\/em>, we write <em>B<\/em>\u00a0\u2286\u00a0<em>A<\/em><\/p>\n<p>A <strong>proper subset<\/strong> is a subset that is not identical to the original set\u2014it excludes at least one element of the original set.<\/p>\n<p>If <em>B<\/em> is a proper subset of <em>A<\/em>, we write <em>B<\/em> \u2282 <em>A<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Consider these three sets:<\/p>\n<p><em>A<\/em> = the set of all even numbers<br \/>\n<em>B<\/em> = {2, 4, 6}<br \/>\n<em>C<\/em> = {2, 3, 4, 6}<\/p>\n<p>Here <em>B<\/em> \u2282 <em>A<\/em> since every element of <em>B<\/em> is also an even number, so is an element of <em>A<\/em>.<\/p>\n<p>More formally, we could say <em>B<\/em> \u2282 <em>A<\/em> since if <em>x <\/em>\u2208\u00a0<em>B<\/em>, then <em>x <\/em>\u2208 <em>A<\/em>.<\/p>\n<p>It is also true that <em>B<\/em> \u2282 <em>C<\/em>.<\/p>\n<p><em>C<\/em> is not a subset of <em>A<\/em>, since C contains an element, 3, that is not contained in <em>A<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125855&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Sets: basics\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5xthPHH4i_A?list=PL7138FAEC01D6F3F3\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose a set contains the plays \u201cMuch Ado About Nothing,\u201d \u201cMacBeth,\u201d and \u201cA Midsummer\u2019s Night Dream.\u201d What is a larger set this might be a subset of?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q42047\">Show Solution<\/span><\/p>\n<div id=\"q42047\" class=\"hidden-answer\" style=\"display: none\">There are many possible answers here. One would be the set of plays by Shakespeare. This is also a subset of the set of all plays ever written. It is also a subset of all British literature.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Consider the\u00a0set\u00a0[latex]A = \\{1, 3, 5\\}[\/latex]. Which of the following sets is [latex]A[\/latex] a subset of?<br \/>\n[latex]X = \\{1, 3, 7, 5\\}[\/latex]<br \/>\n[latex]Y = \\{1, 3 \\}[\/latex]<br \/>\n[latex]Z = \\{1, m, n, 3, 5\\}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3546\">Show Solution<\/span><\/p>\n<div id=\"q3546\" class=\"hidden-answer\" style=\"display: none\"> [latex]X[\/latex] and [latex]Y[\/latex] <\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>Given the\u00a0set: <em>A<\/em> = {<em>a<\/em>, <em>b<\/em>, <em>c<\/em>, <em>d<\/em>}. List all of the subsets of <em>A<br \/>\n<\/em><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q706217\">Show Solution<\/span><em><br \/>\n<\/em><\/p>\n<div id=\"q706217\" class=\"hidden-answer\" style=\"display: none\">{} (or \u00d8), {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d},\u00a0{b,c,d}, {a,b,c,d}<\/p>\n<p>You can see that there are 16 subsets, 15 of which are proper subsets.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Listing the sets is fine if you have only a few elements. However, if we were to list all of the subsets of a set containing many elements, it would be quite tedious. Instead, in the next example we will consider each element of the set separately.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In the previous example, there are four elements. For the first element, <em>a<\/em>, either it\u2019s in the set or it\u2019s not. Thus there are 2 choices for that first element. Similarly, there are two choices for <em>b<\/em>\u2014either it\u2019s in the set or it\u2019s not. Using just those two elements,\u00a0list all the possible subsets of the set {a,b}<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q857946\">Show Solution<\/span><\/p>\n<div id=\"q857946\" class=\"hidden-answer\" style=\"display: none\">\n<p>{}\u2014both elements are not in the set<br \/>\n{<em>a<\/em>}\u2014<em>a<\/em> is in;\u00a0<em>b<\/em> is not in the set<br \/>\n{<em>b<\/em>}\u2014<em>a<\/em> is not in the set;\u00a0<em>b<\/em> is in<br \/>\n{<em>a<\/em>,<em>b<\/em>}\u2014<em>a<\/em> is in; <em>b<\/em> is in<\/p>\n<p>Two choices for <em>a<\/em>\u00a0times the two for <em>b<\/em>\u00a0gives us [latex]2^{2}=4[\/latex] subsets.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now let\u2019s include <em>c,\u00a0<\/em>just for fun. List all the possible subsets of the new set {a,b,c}.<br \/>\nAgain, either <em>c<\/em> is included or it isn\u2019t, which gives us two choices. The outcomes are {}, {<em>a<\/em>}, {<em>b<\/em>}, {<em>c<\/em>}, {<em>a<\/em>,<em>b<\/em>}, {<em>a<\/em>,<em>c<\/em>}, {<em>b<\/em>,<em>c<\/em>}, {<em>a<\/em>,<em>b<\/em>,<em>c<\/em>}. Note that there are [latex]2^{3}=8[\/latex] subsets.<\/p>\n<p>If you include\u00a0four elements, there would be [latex]2^{4}=16[\/latex] subsets. 15 of those subsets are proper, 1 subset, namely {<em>a<\/em>,<em>b<\/em>,<em>c<\/em>,<em>d<\/em>}, is not.<\/p>\n<p>In general, if you have <em>n<\/em> elements in your set, then there are [latex]2^{n}[\/latex] subsets and [latex]2^{n}\u22121[\/latex] proper subsets.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm132343\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=132343&theme=oea&iframe_resize_id=ohm132343&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Union, Intersection, and Complement<\/h2>\n<p>Commonly, sets interact. For example, you and a new roommate decide to have a house party, and you both invite your circle of friends. At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets.<\/p>\n<div class=\"textbox\">\n<h3>Union, Intersection, and Complement<\/h3>\n<p>The <strong>union<\/strong> of two sets contains all the elements contained in either set (or both sets).\u00a0The union is notated <em>A <\/em>\u22c3<em> B.\u00a0<\/em>More formally, <em>x <\/em>\u220a <em>A <\/em>\u22c3 <em>B<\/em> if <em>x <\/em>\u2208 <em>A<\/em> or <em>x <\/em>\u2208 <em>B<\/em> (or both)<\/p>\n<p>The <strong>intersection <\/strong>of two sets contains only the elements that are in both sets.\u00a0The intersection is notated <em>A <\/em>\u22c2<em> B.\u00a0<\/em>More formally, <em>x <\/em>\u2208 <em>A <\/em>\u22c2 <em>B<\/em> if <em>x <\/em>\u2208 <em>A<\/em> and <em>x <\/em>\u2208 <em>B.<\/em><\/p>\n<p>The <strong>complement<\/strong> of a set <em>A<\/em> contains everything that is <em>not<\/em> in the set <em>A<\/em>.\u00a0The complement is notated <em>A\u2019<\/em>, or <em>A<sup>c<\/sup>, or sometimes ~<em>A<\/em>.<\/em><\/p>\n<p>A <strong>universal set<\/strong> is a set that contains all the elements we are interested in. This would have to be defined by the context.<\/p>\n<p>A complement is relative to the universal set, so\u00a0<em>A<sup>c<\/sup><\/em><em>\u00a0<\/em>contains all the elements in the universal set that are not in <em>A<\/em>.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<ol>\n<li>If we were discussing searching for books, the universal set might be all the books in the library.<\/li>\n<li>If we were grouping your Facebook friends, the universal set would be all your Facebook friends.<\/li>\n<li>If you were working with sets of numbers, the universal set might be all whole numbers, all integers, or all real numbers<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose the universal set is <em>U<\/em> = all whole numbers from 1 to 9. If <em>A<\/em> = {1, 2, 4}, then\u00a0<em>A<sup>c<\/sup> <\/em>= {3, 5, 6, 7, 8, 9}.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110368&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Consider the sets:<\/p>\n<p><em>A<\/em> = {red, green, blue}<br \/>\n<em>B<\/em> = {red, yellow, orange}<br \/>\n<em>C<\/em> = {red, orange, yellow, green, blue, purple}<\/p>\n<p>Find the following:<\/p>\n<ol>\n<li>Find <em>A <\/em>\u22c3<em> B<\/em><\/li>\n<li>Find <em>A <\/em>\u22c2<em> B<\/em><\/li>\n<li>Find <em>A<sup>c<\/sup><\/em>\u22c2<em> C<\/em><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q691926\">Show Solution<\/span><\/p>\n<div id=\"q691926\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The union contains all the elements in either set: <em>A <\/em>\u22c3<em> B<\/em> = {red, green, blue, yellow, orange}\u00a0Notice we only list red once.<\/li>\n<li>The intersection contains all the elements in both sets: <em>A <\/em>\u22c2<em> B<\/em> = {red}<\/li>\n<li>Here we\u2019re looking for all the elements that are <em>not<\/em> in set <em>A<\/em> and are also in <em>C<\/em>.\u00a0<em>A<sup>c<\/sup> <\/em>\u22c2<em> C<\/em> = {orange, yellow, purple}<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125865&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"450\"><\/iframe><\/p>\n<\/div>\n<p>Notice that in the example above, it would be hard to just ask for <em>A<sup>c<\/sup><\/em>, since everything from the color fuchsia to puppies and peanut butter are included in the complement of the set. For this reason, complements are usually only used with intersections, or when we have a universal set in place.<\/p>\n<p>As we saw earlier with the expression\u00a0<em>A<sup>c<\/sup><\/em><em>\u00a0<\/em>\u22c2<em> C<\/em>, set operations can be grouped together. Grouping symbols can be used like they are with arithmetic \u2013 to force an order of operations.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose <em>H<\/em> = {cat, dog, rabbit, mouse}, <em>F<\/em> = {dog, cow, duck, pig, rabbit}, and\u00a0<em>W<\/em> = {duck, rabbit, deer, frog, mouse}<\/p>\n<ol>\n<li>Find (<em>H <\/em>\u22c2<em> F<\/em>) \u22c3<em> W<\/em><\/li>\n<li>Find <em>H <\/em>\u22c2 (<em>F<\/em> \u22c3<em> W<\/em>)<\/li>\n<li>Find (<em>H <\/em>\u22c2<em> F<\/em>)<em><sup>c<\/sup><\/em> \u22c2<em> W<\/em><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444204\">Show Solution<\/span><\/p>\n<div id=\"q444204\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>We start with the intersection: <em>H <\/em>\u22c2<em> F<\/em> = {dog, rabbit}.\u00a0Now we union that result with <em>W<\/em>: (<em>H <\/em>\u22c2<em> F<\/em>) \u22c3<em> W<\/em> = {dog, duck, rabbit, deer, frog, mouse}<\/li>\n<li>We start with the union: <em>F<\/em> \u22c3<em> W<\/em> = {dog, cow, rabbit, duck, pig, deer, frog, mouse}.\u00a0Now we intersect that result with <em>H<\/em>: <em>H <\/em>\u22c2 (<em>F<\/em> \u22c3<em> W<\/em>) = {dog, rabbit, mouse}<\/li>\n<li>We start with the intersection: <em>H <\/em>\u22c2<em> F<\/em> = {dog, rabbit}.\u00a0Now we want to find the elements of <em>W<\/em> that are <em>not<\/em> in <em>H <\/em>\u22c2<em> F.\u00a0<\/em>(<em>H <\/em>\u22c2<em> F)<sup>c<\/sup><\/em>\u00a0\u22c2<em> W<\/em> = {duck, deer, frog, mouse}<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3>Venn Diagrams<\/h3>\n<p>To visualize the interaction of sets, John Venn in 1880 thought to use overlapping circles, building on a similar idea used by Leonhard Euler in the 18th century. These illustrations now called <strong>Venn Diagrams<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Venn Diagram<\/h3>\n<p>A Venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Overlapping areas indicate elements common to both sets.<\/p>\n<p>Basic Venn diagrams can illustrate the interaction of two or three sets.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Create Venn diagrams to illustrate <em>A <\/em>\u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>, and <em>Ac <\/em>\u22c2<em> B<\/em><\/p>\n<p><em>A <\/em>\u22c3<em> B<\/em> contains all elements in <em>either<\/em> set.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q252649\">Show Solution<\/span><\/p>\n<div id=\"q252649\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220329\/ab1.png\" alt=\"ab1\" width=\"179\" height=\"125\" \/><\/p>\n<p><em>A <\/em>\u22c2<em> B<\/em> contains only those elements in both sets \u2013 in the overlap of the circles.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220349\/ab2.png\" alt=\"ab2\" width=\"180\" height=\"125\" \/><\/p>\n<p><em>Ac <\/em>will contain all elements <em>not<\/em> in the set A. <em>A<sup>c <\/sup><\/em>\u22c2<em> B<\/em> will contain the elements in set <em>B<\/em> that are not in set <em>A<\/em>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220403\/ab3.png\" alt=\"ab3\" width=\"180\" height=\"125\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use a Venn diagram to illustrate (<em>H <\/em>\u22c2<em> F<\/em>)<sup><em>c<\/em><\/sup> \u22c2<em> W<br \/>\n<\/em><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q311976\">Show Solution<\/span><\/p>\n<div id=\"q311976\" class=\"hidden-answer\" style=\"display: none\">\n<p>We\u2019ll start by identifying everything in the set <em>H <\/em>\u22c2<em> F<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-452\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220617\/hfw1.png\" alt=\"hfw1\" width=\"180\" height=\"152\" \/><\/p>\n<p>Now, (<em>H <\/em>\u22c2<em> F<\/em>)<em>c<\/em> \u22c2<em> W<\/em> will contain everything <em>not<\/em> in the set identified above that is also in set <em>W<\/em>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220632\/hfw2.png\" alt=\"hfw2\" width=\"180\" height=\"156\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Sets: drawing a Venn diagram\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CPeeOUldZ6M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Create an expression to represent the outlined part of the Venn diagram shown.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12220943\/hfw3.png\" alt=\"hfw3\" width=\"180\" height=\"157\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q828282\">Show Solution<\/span><\/p>\n<div id=\"q828282\" class=\"hidden-answer\" style=\"display: none\">\n<p>The elements in the outlined set <em>are<\/em> in sets <em>H<\/em> and <em>F<\/em>, but are not in set <em>W<\/em>. So we could represent this set as <em>H <\/em>\u22c2<em> F<\/em> \u22c2<em> W<sup>c<\/sup><\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Create an expression to represent the outlined portion of the Venn diagram shown.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-455\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12221621\/abc1.png\" alt=\"abc1\" width=\"180\" height=\"157\" \/><br \/>\n<iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6699&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"200\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Sets: writing an expression for a Venn diagram region\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VfC8rhLdYYg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Cardinality<\/h2>\n<p>Often times we are interested in the number of items in a set or subset. This is called the cardinality of the set.<\/p>\n<div class=\"textbox\">\n<h3>Cardinality<\/h3>\n<p>The number of elements in a set is the cardinality of that set.<\/p>\n<p>The cardinality of the set <em>A<\/em> is often notated as |<em>A<\/em>| or n(<em>A<\/em>)<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>Let <em>A<\/em> = {1, 2, 3, 4, 5, 6} and <em>B<\/em> = {2, 4, 6, 8}.<\/p>\n<p>What is the cardinality of <em>B<\/em>? <em>A<\/em> \u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q100844\">Show Solution<\/span><\/p>\n<div id=\"q100844\" class=\"hidden-answer\" style=\"display: none\">\n<p>The cardinality of <em>B<\/em> is 4, since there are 4 elements in the set.<\/p>\n<p>The cardinality of <em>A<\/em> \u22c3<em> B<\/em> is 7, since <em>A<\/em> \u22c3<em> B<\/em> = {1, 2, 3, 4, 5, 6, 8}, which contains 7 elements.<\/p>\n<p>The cardinality of <em>A <\/em>\u22c2<em> B<\/em> is 3, since <em>A <\/em>\u22c2<em> B<\/em> = {2, 4, 6}, which contains 3 elements.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=109846&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>What is the cardinality of <em>P<\/em> = the set of English names for the months of the year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6805\">Show Solution<\/span><\/p>\n<div id=\"q6805\" class=\"hidden-answer\" style=\"display: none\">\n<p>The cardinality of this set is 12, since there are 12 months in the year.<\/p>\n<p>Sometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>A survey asks 200 people \u201cWhat beverage do you drink in the morning\u201d, and offers choices:<\/p>\n<ul>\n<li>Tea only<\/li>\n<li>Coffee only<\/li>\n<li>Both coffee and tea<\/li>\n<\/ul>\n<p>Suppose 20 report tea only, 80 report coffee only, 40 report both.\u00a0\u00a0 How many people drink tea in the morning? How many people drink neither tea or coffee?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q432276\">Show Solution<\/span><\/p>\n<div id=\"q432276\" class=\"hidden-answer\" style=\"display: none\">\n<p>This question can most easily be answered by creating a Venn diagram. We can see that we can find the people who drink tea by adding those who drink only tea to those who drink both: 60 people.<\/p>\n<p>We can also see that those who drink neither are those not contained in the any of the three other groupings, so we can count those by subtracting from the cardinality of the universal set, 200.<\/p>\n<p>200 \u2013 20 \u2013 80 \u2013 40 = 60 people who drink neither.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-458\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222016\/coffeetea.png\" alt=\"coffeetea\" width=\"180\" height=\"124\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125872&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"200\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A survey asks:\u00a0\u00a0 Which online services have you used in the last month:<\/p>\n<ul>\n<li>Twitter<\/li>\n<li>Facebook<\/li>\n<li>Have used both<\/li>\n<\/ul>\n<p>The results show 40% of those surveyed have used Twitter, 70% have used Facebook, and 20% have used both. How many people have used neither Twitter or Facebook?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q778879\">Show Solution<\/span><\/p>\n<div id=\"q778879\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>T<\/em> be the set of all people who have used Twitter, and <em>F<\/em> be the set of all people who have used Facebook. Notice that while the cardinality of <em>F<\/em> is 70% and the cardinality of <em>T<\/em> is 40%, the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> is not simply 70% + 40%, since that would count those who use both services twice. To find the cardinality of <em>F<\/em> \u22c3 <em>T<\/em>, we can add the cardinality of <em>F<\/em> and the cardinality of <em>T<\/em>, then subtract those in intersection that we\u2019ve counted twice. In symbols,<\/p>\n<p>n(<em>F<\/em> \u22c3 <em>T<\/em>) = n(<em>F<\/em>) + n(<em>T<\/em>) \u2013 n(<em>F<\/em> \u22c2 <em>T<\/em>)<\/p>\n<p>n(<em>F<\/em> \u22c3 <em>T<\/em>) = 70% + 40% \u2013 20% = 90%<\/p>\n<p>Now, to find how many people have not used either service, we\u2019re looking for the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> .<\/p>\n<p>Since the universal set contains 100% of people and the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> = 90%, the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> must be the other 10%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The previous example illustrated two important properties called cardinality properties:<\/p>\n<div class=\"textbox\">\n<h3>Cardinality properties<\/h3>\n<ol>\n<li>n(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)<\/li>\n<li>n(<em>Ac<\/em>) = n(<em>U<\/em>) \u2013 n(<em>A<\/em>)<\/li>\n<\/ol>\n<\/div>\n<p>Notice that the first property can also be written in an equivalent form by solving for the cardinality of the intersection:<\/p>\n<p style=\"text-align: center;\">n(<em>A<\/em> \u22c2 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c3 <em>B<\/em>)<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Fifty students were surveyed, and asked if they were taking a social science (SS), humanities (HM) or a natural science (NS) course the next quarter.<\/p>\n<p>21 were taking a SS course\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 26 were taking a HM course<\/p>\n<p>19 were taking a NS course\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9 were taking SS and HM<\/p>\n<p>7 were taking SS and NS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 were taking HM and NS<\/p>\n<p>3 were taking all three\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7 were taking none<\/p>\n<p>How many students are only taking a SS course?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88483\">Show Solution<\/span><\/p>\n<div id=\"q88483\" class=\"hidden-answer\" style=\"display: none\">\n<p>It might help to look at a Venn diagram.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222214\/sshmns.png\" alt=\"sshmns\" width=\"180\" height=\"152\" \/><\/p>\n<p>From the given data, we know that there are<\/p>\n<p>3 students in region <em>e<\/em> and<\/p>\n<p>7 students in region <em>h<\/em>.<\/p>\n<p>Since 7 students were taking a SS and NS course, we know that n(<em>d<\/em>) + n(<em>e<\/em>) = 7. Since we know there are 3 students in region 3, there must be<\/p>\n<p>7 \u2013 3 = 4 students in region <em>d<\/em>.<\/p>\n<p>Similarly, since there are 10 students taking HM and NS, which includes regions <em>e<\/em> and <em>f<\/em>, there must be<\/p>\n<p>10 \u2013 3 = 7 students in region <em>f<\/em>.<\/p>\n<p>Since 9 students were taking SS and HM, there must be 9 \u2013 3 = 6 students in region <em>b<\/em>.<\/p>\n<p>Now, we know that 21 students were taking a SS course. This includes students from regions <em>a, b, d, <\/em>and <em>e<\/em>. Since we know the number of students in all but region <em>a<\/em>, we can determine that 21 \u2013 6 \u2013 4 \u2013 3 = 8 students are in region <em>a<\/em>.<\/p>\n<p>8 students are taking only a SS course.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>One hundred fifty people were surveyed and asked if they believed in UFOs, ghosts, and Bigfoot.<\/p>\n<p>43 believed in UFOs\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 44 believed in ghosts<\/p>\n<p>25 believed in Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 believed in UFOs and ghosts<\/p>\n<p>8 believed in ghosts and Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5 believed in UFOs and Bigfoot<\/p>\n<p>2 believed in all three<\/p>\n<p>How many people surveyed believed in at least one of these things?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q252699\">Show Solution<\/span><\/p>\n<div id=\"q252699\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. There are several answers: The set of all odd numbers less than 10. The set of all odd numbers. The set of all integers. The set of all real numbers.<\/p>\n<p>2. <em>A <\/em>\u22c3<em> C<\/em> = {red, orange, yellow, green, blue purple}<\/p>\n<p><em>Bc <\/em>\u22c2<em> A<\/em> = {green, blue}<\/p>\n<p>3. <em>A <\/em>\u22c3<em> B <\/em>\u22c2<em> Cc<\/em><\/p>\n<p>4. Starting with the intersection of all three circles, we work our way out. Since 10 people believe in UFOs and Ghosts, and 2 believe in all three, that leaves 8 that believe in only UFOs and Ghosts. We work our way out, filling in all the regions. Once we have, we can add up all those regions, getting 91 people in the union of all three sets. This leaves 150 \u2013 91 = 59 who believe in none.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-460\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222338\/ufoghostsbigfoot.png\" alt=\"ufoghostsbigfoot\" width=\"183\" height=\"152\" \/><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom20\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=125868&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Sets: cardinality\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/wErcETeKvrU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-160\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Mathematics for the Liberal Arts I. <strong>Provided by<\/strong>: Extended Learning Institute of Northern Virginia Community College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/eli.nvcc.edu\/\">http:\/\/eli.nvcc.edu\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Learning Objectives and Introduction. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 132343. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Question ID 132343. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 125855. <strong>Authored by<\/strong>: Bohart, Jenifer. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Sets: drawing a Venn diagram. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CPeeOUldZ6M?list=PL7138FAEC01D6F3F3\">https:\/\/youtu.be\/CPeeOUldZ6M?list=PL7138FAEC01D6F3F3<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Sets: drawing a Venn diagram. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CPeeOUldZ6M\">https:\/\/youtu.be\/CPeeOUldZ6M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6699. <strong>Authored by<\/strong>: Morales, Lawrence. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Sets: cardinality. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/wErcETeKvrU\">https:\/\/youtu.be\/wErcETeKvrU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 125872, 109842, 125878. <strong>Authored by<\/strong>:  Bohart,Jenifer, mb Meacham,William. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Mathematics for the Liberal Arts I\",\"author\":\"\",\"organization\":\"Extended Learning Institute of Northern Virginia Community College\",\"url\":\"http:\/\/eli.nvcc.edu\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Learning Objectives and Introduction\",\"author\":\"\",\"organization\":\"Lumen 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