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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.)

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 22E
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Theorem 19 (Commutative Law). If m,n € 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 AUB has size m + n by the above theorem, and BUA has size n + m. Since AUB = BUA, we have m+n=n+m. Theorem 20 (Associative Law). If x, y, z = 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) U C, since it is part of basic set theory.)
Transcribed Image Text:Theorem 19 (Commutative Law). If m,n € 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 AUB has size m + n by the above theorem, and BUA has size n + m. Since AUB = BUA, we have m+n=n+m. Theorem 20 (Associative Law). If x, y, z = 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) U C, since it is part of basic set theory.)
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