/*! 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 2 Verify the equation $$ n(A \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Verify the equation $$ n(A \cup B)=n(A)+n(B) $$ for the given disjoint sets. \(A=\\{x \mid x\) is a whole number between 0 and 4\(\\}\) \(B=\\{x \mid x\) is a negative integer greater than \(-4\\}\)

Short Answer

Expert verified
The elements of set A are \(A=\\{0, 1, 2, 3, 4\\}\), and the elements of set B are \(B=\\{-3, -2, -1\\}\). Since A and B are disjoint, the union of A and B, \(A \cup B = \\{-3, -2, -1, 0, 1, 2, 3, 4\\}\). There are 5 elements in set A, 3 elements in set B, and 8 elements in the union of sets A and B. Since \(n(A \cup B)=n(A)+n(B)\) (8 = 5 + 3), the equation is verified for the given disjoint sets A and B.

Step by step solution

01

List the elements of set A

Set A contains whole numbers between 0 and 4. Hence, \(A=\\{0, 1, 2, 3, 4\\}\).
02

List the elements of set B

Set B contains negative integers greater than -4. Hence, \(B=\\{-3, -2, -1\\}\).
03

Determine the number of elements in A and B

Set A has 5 elements and set B has 3 elements. So, \(n(A)=5\) and \(n(B)=3\).
04

Find the union of sets A and B

Since sets A and B are disjoint (have no common elements), the union of A and B, \(A \cup B\), is the set that contains all elements in A and all elements in B. Thus, \(A \cup B = \\{-3, -2, -1, 0, 1, 2, 3, 4\\}\).
05

Determine the number of elements in \(A \cup B\)

Set \(A \cup B\) contains 8 elements, so \(n(A \cup B)=8\).
06

Verify the equation

The equation states \(n(A \cup B)=n(A)+n(B)\). Since \(n(A)=5\), \(n(B)=3\), and \(n(A \cup B)=8\), the equation holds (8 = 5 + 3). Thus, the equation is verified for the given disjoint sets A and B.

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 91Ó°ÊÓ!

Key Concepts

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

Union of Sets
In set theory, the union of two sets is a fundamental operation that combines all elements from each set. For example, if we have two sets, \(A\) and \(B\), the union, denoted as \(A \cup B\), will include every element that belongs to either \(A\) or \(B\). It essentially collects every unique element from both sets into a single set.

In our exercise, set \(A\) is \(\{0, 1, 2, 3, 4\}\), and set \(B\) is \(\{-3, -2, -1\}\). To find their union, we gather all elements from \(A\) and \(B\) without repeating any numbers, since each number in a set must be unique.

As a result, the union of these two sets is \(A \cup B = \{-3, -2, -1, 0, 1, 2, 3, 4\}\). This union includes every individual number from both sets combined into one comprehensive set without duplicates.
Disjoint Sets
Disjoint sets are sets that have no elements in common. In other words, when you look at two sets and find that their intersection is an empty set, they are considered disjoint.

In our example, set \(A\) includes numbers \(0\) through \(4\), while set \(B\) comprises negative integers \(-3\), \(-2\), and \(-1\). Notice that there is no overlap in these numbers. Since \(A\) and \(B\) have no shared elements, they can be described as disjoint.

This property of being disjoint is crucial when verifying certain equations in set theory, such as determining the count of unique elements in the union of the sets. It simplifies verification because the union of disjoint sets \(A\) and \(B\) equals the sum of their individual cardinalities.
Cardinality
Cardinality is the term used to describe the number of elements within a set. It's a simple yet essential aspect of set theory that helps us analyze and understand collections.

Consider set \(A\), which has elements \(\{0, 1, 2, 3, 4\}\), so its cardinality, or the number of elements, is \(n(A) = 5\). Similarly, set \(B\) has elements \(\{-3, -2, -1\}\), giving it a cardinality of \(n(B) = 3\).

When we determine the union \(A \cup B = \{-3, -2, -1, 0, 1, 2, 3, 4\}\), its cardinality turns out to be \(n(A \cup B) = 8\). Understanding cardinality is fundamental when verifying the cardinality of unions of any sets and evaluating equations related to them.
Verification of Equations
Verification of equations in set theory involves checking the accuracy or correctness of mathematical statements or identities involving sets. It's a way of ensuring that the relationships we assume about sets hold true.

In the problem, we have the equation \(n(A \cup B) = n(A) + n(B)\) to verify. Because sets \(A\) and \(B\) are disjoint, their union's cardinality should indeed equal the sum of their individual cardinalities.

By substituting our earlier findings: \(n(A) = 5\), \(n(B) = 3\), and \(n(A \cup B) = 8\), we can see that \(8 = 5 + 3\) perfectly satisfies the equation. This confirms that our understanding of set operations like the union and properties like cardinality is correctly applied when dealing with disjoint sets. Verification gives us confidence in both our calculations and our theoretical understanding.

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

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$ \begin{array}{lccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

Let \(S\) be a sample space for an experiment. Show that if \(E\) is any event of an experiment, then \(E\) and \(E^{c}\) are mutually exclusive.

In a poll conducted among 2000 college freshmen to ascertain the political views of college students. the accompanying data were obtained. Determine the empirical probability distribution associated with these data. $$ \begin{array}{lccccc} \hline \text { Political Views } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\ \hline \text { Respondents } & 52 & 398 & 1140 & 386 & 24 \\ \hline \end{array} $$ A: Far left B: Liberal C: Middle of the road D: Conservative E: Far right

An opinion poll was conducted among a group of registered voters in a certain state concerning a proposition aimed at limiting state and local taxes. Results of the poll indicated that \(35 \%\) of the voters favored the proposition, \(32 \%\) were against it, and the remaining group were undecided. If the results of the poll are assumed to be representative of the opinions of the state's electorate, what is the probability that a registered voter selected at random from the electorate a. Favors the proposition? b. Is undecided about the proposition?

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}\right\\}\) be the sample space associated with an experiment having the probability distribution shown in the accompanying table. If \(A=\left\\{s_{1}, s_{2}\right\\}\) and \(B=\left\\{s_{1}, s_{3}\right\\}\), find a. \(P(A), P(B)\) b. \(P\left(A^{c}\right), P\left(B^{c}\right)\) c. \(P(A \cap B)\) d. \(P(A \cup B)\) $$ \begin{array}{lc} \hline \text { Outcome } & \text { Probability } \\ \hline s_{1} & \frac{1}{8} \\ \hline s_{2} & \frac{3}{8} \\ \hline s_{3} & \frac{1}{4} \\ \hline s_{4} & \frac{1}{4} \\ \hline \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

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.