Theorem 20 (Associative Law). If x, y, z E N, then (x+y)+z = x+(y+z). Proof. (sketch). This follows from Lemma 17, and the identity AU (BUC) = (AUB) UC. Exercise 9. Write up the above proof. (You do not need to prove the identity AU (BUC) = (AUB) UC, since it is part of basic set theory.)
Theorem 20 (Associative Law). If x, y, z E N, then (x+y)+z = x+(y+z). Proof. (sketch). This follows from Lemma 17, and the identity AU (BUC) = (AUB) UC. Exercise 9. Write up the above proof. (You do not need to prove the identity AU (BUC) = (AUB) UC, since it is part of basic set theory.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Theorem 19 (Commutative Law). If \( m, n \in \mathbb{N} \), then \( m + n = n + m \).
**Proof.** Let \( A \) and \( B \) be disjoint sets such that \( A \) has size \( m \) and \( B \) has size \( n \) (Lemma 16). Now \( A \cup B \) has size \( m + n \) by the above theorem, and \( B \cup A \) has size \( n + m \). Since \( A \cup B = B \cup A \), we have \( m + n = n + m \). □
Theorem 20 (Associative Law). If \( x, y, z \in \mathbb{N} \), then \( (x+y)+z = x+(y+z) \).
**Proof.** (sketch). This follows from Lemma 17, and the identity
\[ A \cup (B \cup C) = (A \cup B) \cup C. \]
□
**Exercise 9.** Write up the above proof. (You do not need to prove the identity \( A \cup (B \cup C) = (A \cup B) \cup C \), since it is part of basic set theory.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee528cef-5ece-4e3b-b391-f983f8531f58%2F0994a297-6ad4-4e33-839c-734c5354f49f%2Fu5l0us_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 19 (Commutative Law). If \( m, n \in \mathbb{N} \), then \( m + n = n + m \).
**Proof.** Let \( A \) and \( B \) be disjoint sets such that \( A \) has size \( m \) and \( B \) has size \( n \) (Lemma 16). Now \( A \cup B \) has size \( m + n \) by the above theorem, and \( B \cup A \) has size \( n + m \). Since \( A \cup B = B \cup A \), we have \( m + n = n + m \). □
Theorem 20 (Associative Law). If \( x, y, z \in \mathbb{N} \), then \( (x+y)+z = x+(y+z) \).
**Proof.** (sketch). This follows from Lemma 17, and the identity
\[ A \cup (B \cup C) = (A \cup B) \cup C. \]
□
**Exercise 9.** Write up the above proof. (You do not need to prove the identity \( A \cup (B \cup C) = (A \cup B) \cup C \), since it is part of basic set theory.)
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