/*! 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 17 Using the laws of set theory, si... [FREE SOLUTION] | 91影视

91影视

Using the laws of set theory, simplify each of the following: a) \(A \cap(B-A)\) b) \((A \cap B) \cup(A \cap B \cap \bar{C} \cap D) \cup(\bar{A} \cap B)\) c) \((A-B) \cup(A \cap B)\) d) \(\bar{A} \cup \bar{B} \cup(A \cap B \cap \bar{C})\)

Short Answer

Expert verified
The simplified expressions for parts (a), (b), (c) and (d) are respectively \( \emptyset \), \( B \), \( A \) and \(\overline{A \cap B}\).

Step by step solution

01

Solving part (a)

The intersection of a set A with the difference of B and A, denoted \(A \cap(B-A)\) simplifies to the empty set, denoted by \( \emptyset \). This is due to the law of difference of sets which states that any common elements between A and A (which is A itself), is removed from set B. The intersection of set A and an empty set is always an empty set.
02

Solving part (b)

Starting with the expression \((A \cap B) \cup(A \cap B \cap \bar{C} \cap D) \cup(\bar{A} \cap B)\). We notice that \(A \cap B \cap \bar{C} \cap D\) is a subset of \(A \cap B\). Therefore the union of a set with one of its subset is equal to the original set itself. This brings the expression to \((A \cap B) \cup(\bar{A} \cap B)\). Now apply the distributive law over the union of sets, resulting in \(B\).
03

Solving part (c)

The expression \((A-B) \cup(A \cap B)\) simplifies to the set A. The law of difference of sets states that any common elements between A and B are removed from A to form (A-B). Thus, (A-B) contains all elements that belong to A but not to B. While, (A 鈭 B) contains all the elements that are in both A and B. These two sets form disjoint parts of set A. Their union therefore constitutes the set A.
04

Solving part (d)

Looking at the expression \(\bar{A} \cup \bar{B} \cup(A \cap B \cap \bar{C})\), we see that the last term, \(A \cap B \cap \bar{C}\) is a subset of \(\bar{B}\). Therefore, we can simplify this to \(\bar{A} \cup \bar{B}\). Using the law of De Morgan, this simplifies further to \(\overline{A \cap B}\). When dealing with set complements, the complement of the intersection of two sets A and B is the same as the union of their complements.

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.

Laws of Set Theory
Understanding the laws of set theory provides a strong foundation for comprehending how sets interact in mathematics. These laws include operations such as union, intersection, and set difference, and they adhere to specific rules that allow us to simplify complex set expressions into more manageable forms. For example, the commutative law tells us that the order of sets in a union or intersection operation does not affect the result, meaning that A 鈭 B = B 鈭 A and A 鈭 B = B 鈭 A.

The associative law allows us to group sets differently without changing the outcome: (A 鈭 B) 鈭 C = A 鈭 (B 鈭 C) and (A 鈭 B) 鈭 C = A 鈭 (B 鈭 C). The identity law introduces the concepts of universal set U, which contains all elements, and the empty set , which contains no elements. For any set A, we have A 鈭 U = U and A 鈭 鈭 = 鈭.

An integral part of these laws is understanding how to handle the set difference and complements. The difference, A - B, includes all elements in A that are not in B, while the complement, denoted 卢础 or 鈭丄, comprises all elements not in A with respect to the universal set U. These laws are crucial in simplifying and analyzing set expressions effectively.
Set Difference
The concept of set difference is a fundamental part of set theory. It's the set of elements that belong to one set and not another. Mathematically, for sets A and B, the set difference A - B is defined as the set containing all elements of A that are not in B. Visually, if you imagine two overlapping circles in a Venn diagram, the set difference would correspond to the part of one circle that does not overlap with the other.

When simplifying set expressions involving differences, it's important to realize that the difference operation can often be used to reduce the complexity of a set expression. In the provided exercise, we can see this when we simplify (A - B) 鈭 (A 鈭 B) to just A, because (A - B) contains all the elements of A not in B, and (A 鈭 B) contains all elements in both A and B. Together, they cover all elements in A, thus simplifying to A by itself.
Distributive Law over Sets
In set theory, the distributive law is similar to that in arithmetic and algebra but applied to sets. This law shows us how the operations of union and intersection can be distributed over each other. Specifically, it tells us that A 鈭 (B 鈭 C) = (A 鈭 B) 鈭 (A 鈭 C) and A 鈭 (B 鈭 C) = (A 鈭 B) 鈭 (A 鈭 C). Put simply, when sets are being combined, we can 'distribute' one operation over another, much like multiplying over addition in algebra.

Using the distributive law, we can often simplify complex expressions or prove set equalities. For instance, in the exercise, (A 鈭 B) 鈭 (卢础 鈭 B) can be distributed and then simplified further to simply B, streamlining the original set expression. Recognizing when and how to apply the distributive law is a skill that can greatly aid in understanding and manipulating set expressions in more advanced mathematics.
De Morgan's Laws
Another pivotal concept in set theory is encapsulated in De Morgan's laws. These laws provide a relationship between the union and intersection through the use of complements. Specifically, De Morgan's laws state that the complement of the union of two sets is equal to the intersection of their complements, and vice versa. Mathematically, we express these as 卢(A 鈭 B) = 卢础 鈭 卢B and 卢(A 鈭 B) = 卢础 鈭 卢B.

De Morgan's laws are especially useful when we want to simplify expressions with complements, like in part (d) of our exercise, where 卢础 鈭 卢B 鈭 (A 鈭 B 鈭 卢C) eventually simplifies to 卢(A 鈭 B). Understanding these laws is essential for anyone studying logic, computer science, or mathematics, as they frequently appear when working with set operations and boolean algebra.

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

The binary string 101101 , where the string is unchanged upon reversing order, is called a palindrome (of length 6\()\). Suppose a binary string of length 6 is randomly generated, with 0 , 1 equally likely for each of the six positions in the string. What is the probability the string is a palindrome if the first and sixth bits (a) are both \(1 ;\) (b) are the same?

Devon has a bag containing 22 poker chips - eight red, eight white, and six blue. Aileen reaches in and withdraws three of the chips, without replacement. Find the probability that Aileen has selected (a) no blue chips; (b) one chip of each color; or (c) at least two red chips.

A coin is loaded so that \(\operatorname{Pr}(\mathrm{H})=2 / 3\) and \(\operatorname{Pr}(\mathrm{T})=1 / 3\). Todd tosses this coin twice. Let \(A, B\) be the events A: The first toss is a tail. B: Both tosses are the same. Are \(A, B\) independent?

Urn 1 contains 14 envelopes (of the same size) - six each contain $$\$ 1$$ and the other eight each contain $$\$ 5$$. Urn 2 contains eight envelopes (of the same size as those in urn 1) - three each contain $$\$ 1$$ and the other five each contain $$\$ 5$$. Three envelopes are randomly selected from urn 1 and transferred to urn 2. If Carmen now draws one envelope from urn 2 , what is the probability her selection contains $$\$ 1$$ ?

A computer dealer finds that the number of laptop computers her dealership sells each day is a random variable \(X\) where the probability distribution for \(X\) is given by $$ \operatorname{Pr}(X=x)= \begin{cases}\frac{c x^{2}}{x !}, & x=1,2,3,4,5 \\ 0, & \text { otherwise }\end{cases} $$ where \(c\) is a constant. Determine (a) the value of \(c\); (b) \(\operatorname{Pr}(X \geq 3)\); (c) \(\operatorname{Pr}(X=4 \mid X \geq 3)\); (d) \(E(X)\); and (e) \(\operatorname{Var}(X)\).

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.