Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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.)](https://content.bartleby.com/qna-images/question/ee528cef-5ece-4e3b-b391-f983f8531f58/0994a297-6ad4-4e33-839c-734c5354f49f/u5l0us_processed.jpeg)
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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