/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Argue that \(E=E F \cup E F^{c},... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 15

Argue that \(E=E F \cup E F^{c}, E \cup F=E \cup F E^{c}\).

Short Answer

Expert verified
We can prove that \(E = EF \cup EF^c\) and \(E \cup F = E \cup FE^c\) using set identities and logical equivalence. To prove \(E = EF \cup EF^c\), we need to show that \(E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\). For any element \(x \in E\), it can either belong to \(F\) or \(F^c\), resulting in \(x \in EF \cup EF^c\); hence, \(E \subseteq EF \cup EF^c\). Conversely, since elements of \(EF\) and \(EF^c\) belong to \(E\), we have \(EF \cup EF^c \subseteq E\). Thus, \(E = EF \cup EF^c\). Now, to prove \(E \cup F = E \cup FE^c\), we need to show that \(E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\). If \(x \in E \cup F\), then \(x \in E\) or \(x \in F\), and we can conclude that \(x \in E \cup FE^c\); hence, \(E \cup F \subseteq E \cup FE^c\). Similarly, if \(x \in E \cup FE^c\), then \(x \in E\) or \(x \in FE^c\), resulting in \(x \in E \cup F\); thus, \(E \cup FE^c \subseteq E \cup F\). Consequently, \(E \cup F = E \cup FE^c\).

Step by step solution

01

Prove that \( E \subseteq EF \cup EF^c\)

To demonstrate that \(E \subseteq EF \cup EF^c\), we have to show that every element of \(E\) also belongs to \(EF \cup EF^c\). For any element \(x\) in \(E\), since \(x \in E\), it can either belong to \(F\) or \(F^c\). If \(x \in F\), then \(x \in EF\), and if \(x \in F^c\), then \(x \in EF^c\). In both cases, \(x \in EF \cup EF^c\), which means that \( E \subseteq EF \cup EF^c\).
02

Prove that \(EF \cup EF^c \subseteq E\)

To demonstrate that \(EF \cup EF^c \subseteq E\), we have to show that every element, \(x\), in the union also belongs to \(E\). If \(x \in EF\), then clearly \(x \in E\), since \(EF\) is the intersection of the sets \(E\) and \(F\). Similarly, if \(x \in EF^c\), then \(x\) must belong to \(E\), since \(EF^c\) is the intersection of the sets \(E\) and \(F^c\). Therefore, \(EF \cup EF^c \subseteq E\). Since we have proven both that \( E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\), we can conclude that \(E = EF \cup EF^c\). Now let's prove the second statement: \(E \cup F = E \cup FE^c\).
03

Prove that \(E \cup F \subseteq E \cup FE^c\)

To show that \(E \cup F \subseteq E \cup FE^c\), we need to demonstrate that every element of the left-hand side belongs to the right-hand side as well. Let \(x\) be an element in \(E \cup F\), meaning \(x \in E\) or \(x \in F\). If \(x \in E\), then of course \(x \in E \cup FE^c\). If \(x \in F\), then \(x\) can either belong to \(E\) or \(E^c\). If \(x \in E\), then \(x \in E \cup FE^c\), and if \(x \in E^c\), then \(x \in FE^c \subseteq E \cup FE^c\). Therefore, \(E \cup F \subseteq E \cup FE^c\).
04

Prove that \(E \cup FE^c \subseteq E \cup F\)

To show that \(E \cup FE^c \subseteq E \cup F\), we need to demonstrate that every element of the left-hand side also belongs to the right-hand side. Let \(x\) be an element in \(E \cup FE^c\), meaning \(x \in E\) or \(x \in FE^c\). If \(x \in E\), then of course \(x \in E \cup F\). If \(x \in FE^c\), then \(x\) is in both \(F\) and \(E^c\), meaning \(x \in F\), and therefore \(x \in E \cup F\). Hence, \(E \cup FE^c \subseteq E \cup F\). Since we have proven that \( E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\), we can conclude that \(E \cup F = E \cup FE^c\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Union of Sets
The Union of Sets is a fundamental concept in set theory, which combines all elements from two or more sets. The union of sets allows us to create a new set that contains every element that appears in any of the original sets. For instance, if we have two sets \(A\) and \(B\), their union, denoted by \(A \cup B\), contains every element that is in \(A\) or \(B\), or in both. This operation is inclusive, meaning that it retains duplicate elements only once in the resulting set.
  • The notation \(A \cup B\) represents all the elements from both set \(A\) and set \(B\).
  • If \(C = \{1, 2, 3\}\) and \(D = \{3, 4, 5\}\), then \(C \cup D = \{1, 2, 3, 4, 5\}\).
The union operation can be easily visualized using a Venn diagram, where the overlapping and separate regions of circles represent the combined and distinct parts of the sets. In our exercise, proving \(E \cup F = E \cup FE^c\) shows how the union operation can still include all necessary elements by distributing across intersections, ensuring that no element from \(F\) is left out, even if it's intersected by the complement of another set.
Intersection of Sets
Intersection of Sets is another key operation in set theory. This process involves identifying and gathering all elements that are common to all sets being considered. When we intersect two sets, let's say \(A\) and \(B\), what we get is a new set containing every element that appears in both \(A\) and \(B\) simultaneously. This operation is useful when we need to find items that are shared between groups.
  • The notation \(A \cap B\) stands for the intersection of sets \(A\) and \(B\).
  • For example, if \(A = \{1, 2, 3\}\) and \(B = \{2, 3, 4\}\), then \(A \cap B = \{2, 3\}\).
In our exercise, the intersection plays a role in denoting parts of set \(E\) that connect with \(F\) or its complement \(F^c\). We use the intersection to prove statements like \(EF \cup EF^c = E\), showing that all components of \(E\) can be accounted for through their connection or disconnection with \(F\). It鈥檚 a powerful way to analyze overlap and separation among sets.
Complement of a Set
The Complement of a Set refers to elements that are not in a particular set when considering a universal set, which contains all possible elements under discussion. For a given set \(A\), its complement, denoted as \(A^c\), is a collection of everything in the universal set that is not in \(A\). This operation is particularly useful in distinguishing or filtering out components relative to a larger context.
  • The complement \(A^c\) includes every element not in \(A\), based on a larger universal set \(U\).
  • If \(U = \{1, 2, 3, 4, 5\}\) and \(A = \{1, 2\}\), then \(A^c = \{3, 4, 5\}\).
In the provided exercise, understanding the role of \(F^c\) is crucial. It's part of the process that allows us to divide \(E\) into components that align either with \(F\) or with its complement, which helps in proving statements like \(E = EF \cup EF^c\) by showing that no element in \(E\) is left unaddressed. This approach underscores how complements can offer a directional view of a set鈥檚 coverage relative to all possibilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is a fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now, what is the probability that it is a fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now, what is the probability that it is a fair coin?

A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\) as follows: \(E_{1}=\\{\) the first pile has exactly 1 ace \(\\}\), \(E_{2}=\\{\) the second pile has exactly 1 ace \(\\}\), \(E_{3}=\\{\) the third pile has exactly 1 ace \(\\}\), \(E_{4}=\\{\) the fourth pile has exactly 1 ace \(\\}\) Use Exercise 23 to find \(P\left(E_{1} E_{2} E_{3} E_{4}\right)\), the probability that each pile has an ace.

Suppose that 5 percent of men and \(0.25\) percent of women are color-blind. A colorblind person is chosen at random. What is the probability of this person being male? Assume that there is an equal number of males and females.

Stores \(A, B\), and \(C\) have 50,75, and 100 employees, and, respectively, 50,60, and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns and this is a woman. What is the probability that she works in-store \(C\)?

If two fair dice are tossed, what is the probability that the sum is \(i, i=2,3, \ldots, 12 ?\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.