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Let \(P, Q, R,\) and \(S\) be subsets of a universal set \(U .\) Assume that \((P-Q) \subseteq(R \cap S)\) (a) Complete the following sentence: For each \(x \in U,\) if \(x \in(P-Q),\) then \(\ldots\) (b) Write a useful negation of the statement in Part (9a). (c) Write the contrapositive of the statement in Part (9a).

Short Answer

Expert verified
(a) For each \(x \in U\), if \(x \in (P-Q)\), then \(x \in (R \cap S)\). (b) For each \(x \in U\), \(x \notin (P-Q)\) and \(x \notin (R \cap S)\). (c) For each \(x \in U\), if \(x \notin (R \cap S)\), then \(x \notin (P-Q)\).

Step by step solution

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1. Understanding the problem

Let's first understand the expressions \((P-Q)\) and \((R \cap S)\). Here, \((P-Q)\) represents the set of elements that belong to set P but not in set Q, while \((R \cap S)\) represents the set of elements that belong to both sets R and S. The given information tells us that \((P-Q) \subseteq (R \cap S)\), which means that every element in the set \((P-Q)\) belongs to the set \((R \cap S)\).
02

2. Complete the sentence (a)

To complete the sentence, we need to determine what happens when \(x \in (P-Q)\). According to the given information, if \(x \in (P-Q)\), then it must also belong to \((R \cap S)\). So, we can complete the sentence as follows: For each \(x \in U\), if \(x \in (P-Q)\), then \(x \in (R \cap S)\).
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3. Negate the statement (b)

Negating a statement involves changing "if ..., then ..." to "not ... and ...". So, negating the statement in Part a would be: For each \(x \in U\), \(x \notin (P-Q)\) and \(x \notin (R \cap S)\).
04

4. Contrapositive of the statement (a) (c)

To find the contrapositive of a statement, we reverse and negate both the hypothesis and conclusion. Thus, the contrapositive of the statement in Part (a) would be: For each \(x \in U\), if \(x \notin (R \cap S)\), then \(x \notin (P-Q)\). In conclusion, to analyze the relation \((P-Q) \subseteq (R \cap S)\), we found that for each \(x \in U\), if \(x \in (P-Q)\), then \(x \in (R \cap S)\) (Part a). We also wrote the negation and the contrapositive of this statement as related information (Parts b and c).

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Key Concepts

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

Set Theory

Set theory is the foundation of modern mathematics. It deals with the study of sets, which are collections of objects. Sets are usually denoted by capital letters and the objects within them, known as elements, can be anything such as numbers, letters, or other sets. In the context of the given problem, we learned that difference and intersection are two fundamental operations in set theory.

The difference between two sets, denoted as \(P-Q\), includes all the elements that are in set \(P\) but not in \(Q\). Conversely, the intersection of two sets, represented by \(R \cap S\), contains the elements that both sets \(R\) and \(S\) share. The exercise demonstrates a subset relationship, denoted by \subseteq\, which means that all elements of the first set are also elements of the second

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Proof Writing

Proof writing is an essential skill in mathematics that allows one to convincingly argue the truth of a statement. In proofs, statements are derived logically from axioms, definitions, and previously established theorems. Good proof writing involves clear logic, precise language, and an organized structure. The exercise involved deducing the logical conclusion that if an element \(x\) belongs to the set difference \(P-Q\), it must also belong to the intersection \(R \cap S\), directly reflecting the original subset relation given in the problem.

  • Clarity: Each step of the proof should follow logically from the previous one.
  • Conciseness: Proofs should be as short as possible while conveying all necessary information.
  • Completeness: All parts of the proof should be accounted for, including handling of all possible cases.

Proofs are not just about showing that something is true, they are about understanding why it is true. The improvement advice for the exercise stresses this aspect, encouraging students not only to follow the steps but also to grasp the underlying logic and principles.

Contrapositive Statement

Understanding the contrapositive is critical in logical reasoning and proof writing. The contrapositive of an implication is a type of logical equivalence that, when shown to be true, also proves the original statement. For an implication of the form 'If P, then Q,' the contrapositive would be 'If not Q, then not P.' It reverses and negates both the hypothesis and conclusion of the original statement. In the exercise, the statement 'For each \(x \in U\), if \(x \in (P-Q)\), then \(x \in (R \cap S)\)' is transformed to its contrapositive: 'For each \(x \in U\), if \(x otin (R \cap S)\), then \(x otin (P-Q)\).'

The value of the contrapositive lies in its equivalence to the original statement; proving the contrapositive is often easier and helps prove the initial statement. This exercise guides learners through this process, enhancing their problem-solving toolkit and understanding of logical structures in mathematics.

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Most popular questions from this chapter

(A Set Theoretic Definition of an Ordered Pair) In elementary mathematics, the notion of an ordered pair introduced at the beginning of this section will suffice. However, if we are interested in a formal development of the Cartesian product of two sets, we need a more precise definition of ordered pair. Following is one way to do this in terms of sets. This definition is credited to Kazimierz Kuratowski \((1896-1980)\). Kuratowski was a famous Polish mathematician whose main work was in the areas of topology and set theory. He was appointed the Director of the Polish Academy of Sciences and served in that position for 19 years. Let \(x\) be an element of the set \(A\), and let \(y\) be an element of the set \(B\). The ordered pair \((x, y)\) is defined to be the set \(\\{\\{x\\},\\{x, y\\}\\}\). That is, $$ (x, y)=\\{\\{x\\},\\{x, y\\}\\} $$ (a) Explain how this definition allows us to distinguish between the ordered pairs (3,5) and (5,3) . (b) Let \(A\) and \(B\) be sets and let \(a, c \in A\) and \(b, d \in B\). Use this definition of an ordered pair and the concept of set equality to prove that \((a, b)=\) \((c, d)\) if and only if \(a=c\) and \(b=d\). An ordered triple can be thought of as a single triple of objects, denoted by \((a, b, c),\) with an implied order. This means that in order for two ordered triples to be equal, they must contain exactly the same objects in the same order. That is, \((a, b, c)=(p, q, r)\) if and only if \(a=p, b=q\) and \(c=r\).

Prove the following proposition: For all sets \(A, B,\) and \(C\) that are subsets of some universal set, if \(A \cap B=A \cap C\) and \(A^{c} \cap B=A^{c} \cap C,\) then \(B=C\).

We have used the choose-an-element method to prove results about sets. This method, however, is a general proof technique and can be used in settings other than set theory. It is often used whenever we encounter a universal quantifier in a statement in the backward process. Consider the following proposition. Let \(a, b\), and t be integers with \(t \neq 0 .\) If t divides a and t divides \(b\), then for all integers \(x\) and \(y\), \(t\) divides \((a x+b y) .\) (a) Whenever we encounter a new proposition, it is a good idea to explore the proposition by looking at specific examples. For example, let \(a=20, b=12,\) and \(t=4 .\) In this case, \(t \mid a\) and \(t \mid b .\) In each of the following cases, determine the value of \((a x+b y)\) and determine if \(t\) divides \((a x+b y)\) i. \(x=1, y=1\) iv. \(x=2, y=-3\) ii. \(x=1, y=-1\) v. \(x=-2, y=3\) iii. \(x=2, y=2\) vi. \(x=-2, y=-5\) (b) Repeat Part (18a) with \(a=21, b=-6,\) and \(t=3\). Notice that the conclusion of the conditional statement in this proposition involves the universal quantifier. So in the backward process, we would have \(Q:\) For all integers \(x\) and \(y, t\) divides \(a x+b y\) The "elements" in this sentence are the integers \(x\) and \(y .\) In this case, these integers have no "given property" other than that they are integers. The "something that happens" is that \(t\) divides \((a x+b y)\). This means that in the forward process, we can use the hypothesis of the proposition and choose integers \(x\) and \(y .\) That is, in the forward process, we could have \(P: a, b,\) and \(t\) are integers with \(t \neq 0, t\) divides \(a\) and \(t\) divides \(b\). \(P 1:\) Let \(x \in \mathbb{Z}\) and let \(y \in \mathbb{Z}\) (c) Complete the following proof of Proposition 5.16 . Let \(a, b,\) and \(t\) be integers with \(t \neq 0,\) and assume that \(t\) divides \(a\) and \(t\) divides \(b\). We will prove that for all integers \(x\) and \(y, t\) divides \((a x+b y)\) So let \(x \in \mathbb{Z}\) and let \(y \in \mathbb{Z} .\) Since \(t\) divides \(a,\) there exists an integer \(m\) such that \(\ldots\)

Let \(A=\\{1\\}, B=\\{2\\},\) and \(C=\\{3\\}\). (a) Explain why \(A \times B \neq B \times A\). (b) Explain why \((A \times B) \times C \neq A \times(B \times C)\).

Let \(C=\\{x \in \mathbb{Z} \mid x \equiv 7(\bmod 9)\\}\) and \(D=\\{x \in \mathbb{Z} \mid x \equiv 1(\bmod 3)\\}\). (a) List at least five different elements of the set \(C\) and at least five elements of the set \(D\). (b) Is \(C \subseteq D\) ? Justify your conclusion with a proof or a counterexample. (c) Is \(D \subseteq C\) ? Justify your conclusion with a proof or a counterexample.

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