{"id":234350,"date":"2025-06-14T04:50:56","date_gmt":"2025-06-14T04:50:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234350"},"modified":"2025-06-14T04:50:58","modified_gmt":"2025-06-14T04:50:58","slug":"prove-that-a-intersection-b-whole-complement-is-equals-to-a-complement-union-b-complement","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/prove-that-a-intersection-b-whole-complement-is-equals-to-a-complement-union-b-complement\/","title":{"rendered":"prove that a intersection b whole complement is equals to a complement Union B complement"},"content":{"rendered":"\n

prove that a intersection b whole complement is equals to a complement Union B complement<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Proof: (A\u2229B)c=Ac\u222aBc(A \\cap B)^c = A^c \\cup B^c(A\u2229B)c=Ac\u222aBc<\/strong><\/h3>\n\n\n\n

This is a classic result from set theory known as De Morgan\u2019s Law<\/strong>. It states that the complement of the intersection of two sets is equal to the union of their complements.<\/p>\n\n\n\n

\n\n\n\n

Proof:<\/strong><\/h3>\n\n\n\n

Let x\u2208(A\u2229B)cx \\in (A \\cap B)^cx\u2208(A\u2229B)c.
This means that x\u2209A\u2229Bx \\notin A \\cap Bx\u2208\/A\u2229B.<\/p>\n\n\n\n

By the definition of intersection, x\u2209A\u2229Bx \\notin A \\cap Bx\u2208\/A\u2229B means that x is not in both A and B at the same time<\/strong>.<\/p>\n\n\n\n

So, x\u2209Ax \\notin Ax\u2208\/A or<\/strong> x\u2209Bx \\notin Bx\u2208\/B.<\/p>\n\n\n\n

Therefore, x\u2208Acx \\in A^cx\u2208Ac or x\u2208Bcx \\in B^cx\u2208Bc, which implies x\u2208Ac\u222aBcx \\in A^c \\cup B^cx\u2208Ac\u222aBc.<\/p>\n\n\n\n

Hence,x\u2208(A\u2229B)c\u21d2x\u2208Ac\u222aBcx \\in (A \\cap B)^c \\Rightarrow x \\in A^c \\cup B^cx\u2208(A\u2229B)c\u21d2x\u2208Ac\u222aBc<\/p>\n\n\n\n

This shows that:(A\u2229B)c\u2286Ac\u222aBc(A \\cap B)^c \\subseteq A^c \\cup B^c(A\u2229B)c\u2286Ac\u222aBc<\/p>\n\n\n\n

Now, take x\u2208Ac\u222aBcx \\in A^c \\cup B^cx\u2208Ac\u222aBc.
Then x\u2208Acx \\in A^cx\u2208Ac or x\u2208Bcx \\in B^cx\u2208Bc, which means x\u2209Ax \\notin Ax\u2208\/A or x\u2209Bx \\notin Bx\u2208\/B.<\/p>\n\n\n\n

Thus, x cannot be in both A and B<\/strong>, so x\u2209A\u2229Bx \\notin A \\cap Bx\u2208\/A\u2229B.<\/p>\n\n\n\n

This implies x\u2208(A\u2229B)cx \\in (A \\cap B)^cx\u2208(A\u2229B)c.<\/p>\n\n\n\n

Hence,x\u2208Ac\u222aBc\u21d2x\u2208(A\u2229B)cx \\in A^c \\cup B^c \\Rightarrow x \\in (A \\cap B)^cx\u2208Ac\u222aBc\u21d2x\u2208(A\u2229B)c<\/p>\n\n\n\n

So,Ac\u222aBc\u2286(A\u2229B)cA^c \\cup B^c \\subseteq (A \\cap B)^cAc\u222aBc\u2286(A\u2229B)c<\/p>\n\n\n\n

Since both sets are subsets of each other, we conclude:(A\u2229B)c=Ac\u222aBc(A \\cap B)^c = A^c \\cup B^c(A\u2229B)c=Ac\u222aBc<\/p>\n\n\n\n

\n\n\n\n

Explanation (300 Words):<\/strong><\/h3>\n\n\n\n

In set theory, De Morgan’s Laws are powerful tools that describe how complementation interacts with union and intersection. The law we are proving here says that the complement of the intersection of two sets is equal to the union of their complements:(A\u2229B)c=Ac\u222aBc(A \\cap B)^c = A^c \\cup B^c(A\u2229B)c=Ac\u222aBc<\/p>\n\n\n\n

To understand this, consider what it means for an element to belong to the complement of the intersection (A\u2229B)c(A \\cap B)^c(A\u2229B)c. If a value lies in this set, then it is not<\/strong> found in both A and B. That is, it either does not<\/strong> belong to A, or it does not<\/strong> belong to B\u2014or possibly neither.<\/p>\n\n\n\n

If the element is not in A, then it belongs to AcA^cAc.
If the element is not in B, then it belongs to BcB^cBc.
If either of these is true, then the element lies in the union Ac\u222aBcA^c \\cup B^cAc\u222aBc.<\/p>\n\n\n\n

Conversely, if an element is in Ac\u222aBcA^c \\cup B^cAc\u222aBc, it is not in A or not in B, which means it cannot be in both simultaneously. So it must lie in (A\u2229B)c(A \\cap B)^c(A\u2229B)c.<\/p>\n\n\n\n

Thus, each side of the equation captures exactly the same elements. This fundamental identity is useful in simplifying expressions in probability, logic, and computer science. It helps transform complex set expressions into more manageable forms, especially when dealing with complements.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

prove that a intersection b whole complement is equals to a complement Union B complement The Correct Answer and Explanation is: Proof: (A\u2229B)c=Ac\u222aBc(A \\cap B)^c = A^c \\cup B^c(A\u2229B)c=Ac\u222aBc This is a classic result from set theory known as De Morgan\u2019s Law. It states that the complement of the intersection of two sets is equal […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-234350","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234350","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=234350"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234350\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234350"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234350"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234350"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}