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Verify the equation $$ n(A \cup B)=n(A)+n(B) $$ for the given disjoint sets. $$ A=\\{a, e, i, o, u\\} \text { and } B=\\{g, h, k, l, m\\} $$

Short Answer

Expert verified
We are given disjoint sets \(A=\{a, e, i, o, u\}\) and \(B=\{g, h, k, l, m\}\). Finding the cardinality of both sets, we have \(n(A)=5\) and \(n(B)=5\). The union of sets A and B is \(A \cup B = \{a, e, i, g, h, k, l, m, o, u\}\) and its cardinality is \(n(A \cup B) = 10\). Since \(n(A \cup B) = n(A) + n(B) = 10\), the given equation holds true for these disjoint sets.

Step by step solution

01

Finding the cardinality of set A

: Set A contains the elements:\( A = \{a, e, i, o, u\} \). So, the number of elements in set A is: $$ n(A) = 5 $$
02

Finding the cardinality of set B

: Set B contains the elements: \( B = \{g, h, k, l, m\} \). So, the number of elements in set B is: $$ n(B) = 5 $$
03

Finding the union of sets A and B

: Since sets A and B are disjoint (they have no common elements), the union of the two sets will contain all the elements from set A and set B. The union of sets A and B is denoted by \(A \cup B\) and can be written as: $$ A \cup B = \{a, e, i, g, h, k, l, m, o, u\} $$
04

Finding the cardinality of the union of sets A and B

: The set \(A \cup B\) contains 10 elements total. So, the cardinality of the union of sets A and B is: $$ n(A \cup B) = 10 $$
05

Verify the given equation

: Now let's verify if the given equation \(n(A \cup B) = n(A) + n(B)\) holds true. We have: $$ n(A \cup B) = 10 \\ n(A) + n(B) = 5 + 5 = 10 $$ Since both sides of the equation are equal, the given equation holds true for these disjoint sets.

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Key Concepts

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

Cardinality of Sets

The cardinality of a set refers to the number of elements contained within that set. It is a measure of the \'size\' of the set. Cardinality is denoted by the letter \'n\', followed by the set in parentheses, such as in the expression \(n(A)\) for set A. To find the cardinality, you simply count the distinct elements present.


For example, in the provided exercise, we calculated the cardinality of set A, which contains the vowels: \( A = \{a, e, i, o, u\} \), and determined that \(n(A) = 5\). Similarly, we determined the cardinality of set B, with the consonants: \( B = \{g, h, k, l, m\} \), as \(n(B) = 5\). Understanding the cardinality of a set is essential when comparing sets or performing operations such as unions or intersections.

Union of Sets

The union of two or more sets consists of all the distinct elements that appear in any of the sets. Mathematically, we denote the union of sets A and B as \(A \cup B\). If an element is part of either set A or set B, or both, it will be included in the union of these sets.


In the case of the exercise, since sets A and B are disjoint, we know they share no common elements, which simplifies the process of forming their union. Therefore, \(A \cup B\) will encompass all the elements from both A and B without any overlap. As demonstrated in the solution, \(A \cup B = \{a, e, i, g, h, k, l, m, o, u\}\), which includes all elements from both sets combining into a larger set with a total cardinality of ten.

Disjoint Sets

Disjoint sets are sets that do not share any elements, meaning they have no overlap. When two sets are disjoint, their intersection, the set of elements they have in common, is empty. This concept is essential when considering union operations because the union of disjoint sets will simply be the combination of all elements from both sets without duplications.


In our exercise, the sets A (vowels) and B (consonants) are perfect examples of disjoint sets, as there is no letter that is both a vowel and a consonant. Acknowledging that the sets are disjoint confirms that their union will have a cardinality equal to the sum of the individual cardinalities; hence, \(n(A \cup B) = n(A) + n(B)\) as illustrated previously.

Applied Mathematics

Applied mathematics is the application of mathematical methods by different fields such as science, engineering, business, computer science, and industry. In applied mathematics, the set theory concept is often used to solve problems involving data classification, statistics, probability, and more. It aids in understanding the relationship between different data groups or sets, as seen in the textbook exercise with disjoint sets. By calculating the cardinality of sets and their unions, students embody the role of applied mathematicians, handling real-world problems that can be abstracted and processed through mathematical concepts.


Understanding how set theory pertains to applied mathematics gives students a practical framework to apply these skills beyond the classroom, fostering analytical and problem-solving abilities that are invaluable in various professions and in research.

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