Chapter 1: Problem 17
P and Q are subsets of the universal set U with the three sets defined as
follows. \(U=\\{x |-2 \leq x \leq 4\\}, P=\\{x |-2 \leq x \leq 1\\}, Q=\\{x |
0
Short Answer
Expert verified
a) \(\{x | 0 < x \leq 1\}\); b) \(\{x | -2 \leq x < 3\}\); c) \(\{x | 1 < x \leq 4\}\); d) \(\{x | -2 \leq x \leq 0\} \cup \{x | 3 \leq x \leq 4\}\); e) \(\{x | 3 \leq x \leq 4\}\); f) \(\{x | -2 \leq x \leq 0\} \cup \{x | 1 < x \leq 4\}\)."}
Step by step solution
01
Find Intersection of P and Q
To find the intersection \(P \cap Q\), we need to determine the values of \(x\) that are common in both sets \(P\) and \(Q\). \(P = \{x | -2 \leq x \leq 1\}\) and \(Q = \{x | 0 < x < 3\}\). So, \(P \cap Q\) is the set of \(x\) values that satisfy both inequalities: \(0 < x \leq 1\). Hence, \(P \cap Q = \{x | 0 < x \leq 1\}\).
02
Find Union of P and Q
The union \(P \cup Q\) consists of all values in set \(P\) or set \(Q\) or in both. Since \(P = \{x | -2 \leq x \leq 1\}\) and \(Q = \{x | 0 < x < 3\}\), \(P \cup Q\) includes all \(x\) from \(-2\) to \(1\), and from more than \(0\) to less than \(3\). This implies \(P \cup Q = \{x | -2 \leq x < 3\}\).
03
Find Complement of P
The complement \(P'\) consists of all the elements in \(U\) that are not in \(P\). Since \(U = \{x | -2 \leq x \leq 4\}\) and \(P = \{x | -2 \leq x \leq 1\}\), \(P'\) will be \(\{x | 1 < x \leq 4\}\).
04
Find Complement of Q
The complement \(Q'\) includes all elements in \(U\) that aren't in \(Q\). Given \(U = \{x | -2 \leq x \leq 4\}\) and \(Q = \{x | 0 < x < 3\}\), \(Q'\) will be \(\{x | -2 \leq x \leq 0\} \cup \{x | 3 \leq x \leq 4\}\).
05
Find Complement of the Union of P and Q
The complement \((P \cup Q)'\) consists of elements in \(U\) that are not in \(P \cup Q\). From Step 2, we found \(P \cup Q = \{x | -2 \leq x < 3\}\). Thus, \((P \cup Q)'\) is \(\{x | 3 \leq x \leq 4\}\).
06
Find Complement of the Intersection of P and Q
The complement \((P \cap Q)'\) consists of all elements in \(U\) that are not in \(P \cap Q\). From Step 1, \(P \cap Q = \{x | 0 < x \leq 1\}\). Thus, \((P \cap Q)'\) is \(\{x | -2 \leq x \leq 0\} \cup \{x | 1 < x \leq 4\}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Subsets
In set theory, a subset is a set where every element is also contained within another set, known as the superset. If we say a set A is a subset of set B, this means that every element of A is also an element of B. Subsets can be strict or non-strict. A strict subset means A contains fewer elements than B, whereas a non-strict subset could mean A and B are the same set.
- If set P, with elements from \(-2\) to \(1\), is part of a larger set U, which includes elements from \(-2\) to \(4\), then P is a subset of U.
- Set Q, containing elements between \(0\) and less than \(3\), can also be a subset of U as it fits inside the boundaries of U's elements.
Union and Intersection
The union and intersection are fundamental operations in set theory. They help us combine and find commonality among sets.**Union:** The union of two sets, denoted as \(P \cup Q\), consists of elements that are in either set or both. For instance, combining the set P \([-2, 1]\)\ and set Q \(0, 3\) would yield a union that includes elements from \(-2\) to \(3\) without overlapping any numerical values. In this exercise, \(P \cup Q\) results in \(\{x | -2 \leq x < 3\}\).**Intersection:** The intersection of sets, represented as \(P \cap Q\), includes only elements present in both sets. For sets P and Q mentioned, the intersection is the numbers both sets specify, resulting in \(\{x | 0 < x \leq 1\}\), as only these numbers satisfy both conditions of set P and Q.Through these operations, we're able to summarize and manipulate complex sets to see which numbers fit multiple conditions.
Complements
In set theory, the complement of a set contains all the elements that are not present in the specified set but exist in a universal set (U). The notation for the complement is usually indicated by a prime (Einauchacion.
- For set P, where P \(= \{x | -2 \leq x \leq 1\}\), its complement P' would include elements that are part of the universal set U but not in P, resulting in \(\{x | 1 < x \leq 4\}\).
- Similarly, for Q \(= \{x | 0 < x < 3\}\), the complement Q' would be \[\{x | -2 \leq x \leq 0\} \cup \{x | 3 \leq x \leq 4\}\]\. This means all values not in set Q within U.
Universal Set
The universal set, often marked as U, encompasses all possible elements under consideration in a particular problem or discussion. It's essentially the 'universe' of items, including every potential member involved.In this exercise, the universal set U is defined as \(\{x | -2 \leq x \leq 4\}\). This set includes all elements from \(-2\) to \(4\), covering both ends completely.
- It serves as the basis for determining the sets that form subsets, like P and Q.
- In terms of complements, universal sets help identify what elements fall outside of any subset's definition. This is done by contrasting the subset with U.