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Chapter 5: Problem 31

For all sets \(A\) and \(B, A-(A \cap B)=A-B\).

Short Answer

Expert verified
We defined A - (A 鈭� B) and A - B as sets containing elements in A but not in the intersection of A and B, and in A but not in B, respectively. By analyzing elements x from both sets, we showed that if x 鈭� A - (A 鈭� B), then x 鈭� A - B, and if x 鈭� A - B, then x 鈭� A - (A 鈭� B). This implies A - (A 鈭� B) = A - B.

Step by step solution

01

Define A - (A 鈭� B) and A - B

Given two sets A and B, let's define the following notations: 1. A - (A 鈭� B): This represents the set of all elements that are in set A but not in the intersection of A and B. 2. A - B: This represents the set of all elements that are in set A but not in set B.
02

Analyze A - (A 鈭� B)

Let x be an arbitrary element of A - (A 鈭� B). We need to show that x must be an element of A - B. By definition, if x 鈭� A - (A 鈭� B), then x 鈭� A and x 鈭� A 鈭� B. Since x 鈭� A 鈭� B, x cannot be in both A and B simultaneously. However, we know that x 鈭� A, so x must not be in B. This means that x 鈭� A and x 鈭� B, so x 鈭� A - B. Therefore, A - (A 鈭� B) 鈯� A - B.
03

Analyze A - B

Now let x be an arbitrary element of A - B. We need to show that x must be an element of A - (A 鈭� B). By definition, if x 鈭� A - B, then x 鈭� A and x 鈭� B. Since x 鈭� B, x cannot be in both A and B simultaneously. This means that x 鈭� A 鈭� B, even though x 鈭� A. Thus, it follows that x 鈭� A - (A 鈭� B). Therefore, A - B 鈯� A - (A 鈭� B).
04

Conclude the proof

Since we have shown that A - (A 鈭� B) 鈯� A - B and A - B 鈯� A - (A 鈭� B), we can conclude that A - (A 鈭� B) = A - B, which completes the proof.

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Key Concepts

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

Set Operations
Set operations are fundamental tools used in mathematics to compare and manipulate different sets. These operations help us understand the relationship between sets and can be applied in various contexts, like probability, logic, and computer science.

The most common set operations include:
  • Union (\( A \cup B \)) - A set containing all the elements that are in A, in B, or in both.
  • Intersection (\( A \cap B \)) - A set containing only the elements that are in both A and B.
  • Difference (\( A - B \)) - A set containing elements that are in A but not in B.
  • Complement (\( A^c \)) - A set containing all the elements not in A.
By mastering these operations, you can solve various mathematical problems and gain a deeper understanding of how different sets interact with one another. For example, in the exercise above, the difference operation is used to show that two expressions for the set difference are equivalent.
Intersection of Sets
The intersection of sets is one of the key concepts in set theory and represents the set that contains only elements common to all sets involved.

If we talk about two sets, A and B, their intersection is denoted as (\( A \cap B \)). This means that it contains every element that is both in A and in B. For instance, if set A represents all students who play soccer, and set B represents all students who play basketball, then (\( A \cap B \)) includes students who play both soccer and basketball.

Understanding intersecting sets helps simplify expressions and calculations regarding sets. In the provided exercise, we explored \( A - (A \cap B) \) meaning elements in A not in the overlap of A and B. This insight helps us better articulate and prove the relationship between various groupings of elements in overlapping sets.
Complement of Sets
The complement of a set is a way to identify what is not part of a particular set within a universal set. If you have a universal set, such as all the students in a class, and a subset A, say students wearing glasses, the complement of A (\( A^c \)) would be students not wearing glasses.

Formalized, the complement of set A is expressed as (\( A^c \)) and it consists of all elements in the universal set that are not in A.
  • This operation is incredibly useful when dealing with probabilities and logic as it helps us determine not just what's there but what's missing.
  • When combining complements with other operations like intersection and difference, it offers a robust toolset for logical deductions and solving mathematical problems.
While the exercise provided doesn't directly use complements, understanding them is critical as they form the basis for more complex operations involving set differences.

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Most popular questions from this chapter

Prove that for all sets \(A\) and \(B,(A \cap B)^{c}=A^{c} \cup B^{c}\). Use an element argument to prove each statement in 8-17. Assume that all sets are subsets of a universal set \(U\).

Some English adjectives are descriptive of themselves (for instance, the word polysyllabic is polysyllabic) whereas others are not (for instance, the word monosyllabic is not monosyllabic). The word heterological refers to an adjective that does not describe itself. Is heterological heterological? Explain your answer.

Assume that \(B\) is a Boolean algebra with operations \(+\) and . Give the reasons needed to fill in the blanks in the proofs, but do not use any parts of Theorem \(5.3 .2\) unless they have already been proved. You may use any part of the definition of a Boolean algebra and the results of previous exercises, however. For all \(a\) and \(b\) in \(B,(a+b) \cdot a=a\). Proof: Let \(a\) and \(b\) be any elements of \(B\). Then $$ \begin{aligned} (a+b) \cdot a &=a \cdot(a+b) & & \frac{(\mathrm{a})}{(\mathrm{b})} \\ &=a \cdot a+a \cdot b & & \frac{(\mathrm{d})}{(\mathrm{c})} \\ &=a+a \cdot b & & \frac{(\mathrm{d})}{(\mathrm{d})} \\ &=a \cdot 1+a \cdot b & & \\ &=a \cdot(1+b) & & \frac{(\mathrm{e})}{\mathrm{by} \text { exercise } 49} \\ &=a \cdot(b+1) & & \text { (f) } \\ &=a \cdot 1 & & \end{aligned} $$

$$ \text { If } U \text { denotes a universal set, then } U^{c}=\emptyset \text {. } $$

Consider the following set property: For all sets \(A, B\), and \(C,(A-B) \cup(B-C)=(A \cup B)-(B \cap C)\). a. Use an element argument to derive the property. b. Use an algebraic argument to derive the property (by applying properties from Theorem \(5.2 .2\) ). c. Comment on which method you found easier.
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