\n| Types of Sets and their Symbols<\/a><\/td>\n | Standard Sets of Numbers<\/a><\/td>\n | <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nComplement of a Set Examples<\/h3>\nExample 1:<\/strong><\/p>\nIf A = {1, 3, 5, 7, 9, 11}, \u222a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find the complement of A.<\/p>\n Solution:<\/strong><\/p>\nGiven sets are<\/p>\n A = {1, 3, 5, 7, 9, 11}, \u222a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}<\/p>\n Ac<\/sup> = \u222a – A<\/p>\n= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1, 3, 5, 7, 9, 11}<\/p>\n = {2, 4, 6, 8, 10}<\/p>\n Therefore, Ac<\/sup> = {2, 4, 6, 8, 10}<\/p>\nExample 2:<\/strong><\/p>\nIf \u222a = {5, 10, 15, 20, 25, 30, 35}, B = {10, 20, 30, 40, 50, 60}, evaluate the complement of set B.<\/p>\n Solution:<\/strong><\/p>\nGiven sets are<\/p>\n \u222a = {5, 10, 15, 20, 25, 30, 35}, B = {10, 20, 30, 40, 50, 60}<\/p>\n B’ = \u222a – B<\/p>\n B’ means include the elements of the universal set that are not in set B.<\/p>\n B’ = {5, 10, 15, 20, 25, 30, 35} – {10, 20, 30, 40, 50, 60}<\/p>\n = {5, 15, 25, 35}<\/p>\n Therefore, B’ = {5, 15, 25, 35}.<\/p>\n Example 3:<\/strong><\/p>\nIf P = {2, 3, 5, 7, 9}, Q = {1, 4, 5, 8, 10}, \u222a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then prove that (P \u2229 Q)’ = P’ \u222a Q’<\/p>\n Solution:<\/strong><\/p>\nGiven sets are<\/p>\n P = {2, 3, 5, 7, 9}, Q = {1, 4, 5, 8, 10}, \u222a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}<\/p>\n P’ = \u222a – P<\/p>\n = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {2, 3, 5, 7, 9}<\/p>\n = {1, 4, 6, 8, 10}<\/p>\n Q’ = \u222a – Q<\/p>\n = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {1, 4, 5, 8, 10}<\/p>\n = {2, 3, 6, 7, 9}<\/p>\n P \u2229 Q = {2, 3, 5, 7, 9} \u2229 {1, 4, 5, 8, 10}<\/p>\n = {5}<\/p>\n L.H.S = (P \u2229 Q)’<\/p>\n = \u222a – (P \u2229 Q)<\/p>\n = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5}<\/p>\n = {1, 2, 3, 4, 6, 7, 8, 9, 10}<\/p>\n R.H.S = P’ \u222a Q’<\/p>\n = {1, 4, 6, 8, 10} \u222a {2, 3, 6, 7, 9}<\/p>\n = {1, 2, 3, 4, 6, 8, 7, 9, 10}<\/p>\n L.H.S = R.H.S<\/p>\n Hence, proved.<\/p>\n Example 4:<\/strong><\/p>\nIf A = {l, a, d, r}, \u222a = {a, b, c, d, k, l, r, s t, y},\u00a0 then show that A \u222a A’ = A’ \u222a A = \u222a<\/p>\n Solution:<\/strong><\/p>\nGiven sets are<\/p>\n A = {l, a, d, r}, \u222a = {a, b, c, d, k, l, r, s t, y}<\/p>\n To prove that, A \u222a A’ = A’ \u222a A = \u222a<\/p>\n A’ = \u222a – A<\/p>\n = {a, b, c, d, k, l, r, s t, y} – {l, a, d, r}<\/p>\n = {b, c, k, s, t, y}<\/p>\n A \u222a A’ = {l, a, d, r} \u222a {b, c, k, s, t, y}<\/p>\n = {l, a, d, r, b, c, k, s, t, y} = \u222a<\/p>\n A’ \u222a A = {b, c, k, s, t, y} \u222a {l, a, d, r}<\/p>\n = {b, c, k, s, t, y, l, a, d, r} = \u222a<\/p>\n Hence, shown.<\/p>\n FAQs on Set Complement<\/h3>\n1. Which law deals with the complement of a set?<\/strong><\/p>\nThe various laws which deal with the set complement are listed here.<\/p>\n Complement laws (union and intersection) A \u222a A\u2019 = U, A \u2229 A\u2019 = \u2205<\/p>\n Law of double complementation (A’)’ = A<\/p>\n Law of empty and universal set \u2205\u2019 = U, U\u2019 = \u2205<\/p>\n De Morgan’s Complement Law (A\u222aB)\u2019 = A\u2019 \u2229 B\u2019, (A \u2229 B)\u2019 = A\u2019 \u222a B\u2019<\/p>\n 2. How do you solve complements of a set?<\/strong><\/p>\nThe complement of a set is obtained by subtracting the universal set from the given set. The resultant set complement has the set of elements that are in the universal set and not in the given set.<\/p>\n 3. What is the symbol of a complement of a set?<\/strong><\/p>\nThe complement of a set M is represented as M’ or Mc<\/sup>.<\/p>\n4. What is a universal set?<\/strong><\/p>\nA universal set is a set that includes all the elements of other sets, including its own elements. It is represented by ‘U’.<\/p>\n","protected":false},"excerpt":{"rendered":" A set of a pack of well-defined objects. Those objects are called the elements or members of a set. The complement of a set is nothing but the subtraction of a universal set from any of the given set. The set complement contains the non-common elements of the universal set. Get to know more about … Read more<\/a><\/p>\n","protected":false},"author":12,"featured_media":86396,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"https:\/\/bigideasmathanswers.com\/wp-content\/uploads\/2021\/04\/Complement-of-a-Set-1.png","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/85622"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/12"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=85622"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/85622\/revisions"}],"predecessor-version":[{"id":91412,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/85622\/revisions\/91412"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media\/86396"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=85622"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=85622"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=85622"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} |
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