Abstract Algebra Theory and Applications - Computer Science ...

392 HINTS AND SOLUTIONSConversely, let y ∈ f(A 1 ) ∪ f(A 2 ) ⇒ y ∈ f(A 1 ) or f(A 2 ) ⇒ there existsan x ∈ A 1 or there exists an x ∈ A 2 such that f(x) = y ⇒ there exists anx ∈ A 1 ∪ A 2 such that f(x) = y ⇒ f(A 1 ) ∪ f(A 2 ) ⊂ f(A 1 ∪ A 2 ). Hence,f(A 1 ∪ A 2 ) = f(A 1 ) ∪ f(A 2 ).25. (a) Not an equivalence relation. Fails to be symmetric.(c) Not an equivalence relation. Fails to be transitive.28. Let X = N ∪ { √ 2 } and define x ∼ y if x + y ∈ N.Chapter 1. The Integers1. S(1) : [1(1 + 1)(2(1) + 1)]/6 = 1 = 1 2 is true. Assume S(k) : 1 2 + 2 2 +· · · + k 2 = [k(k + 1)(2k + 1)]/6 is true. Then 1 2 + 2 2 + · · · + k 2 + (k + 1) 2 =[k(k + 1)(2k + 1)]/6 + (k + 1) 2 = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6, soS(k + 1) is true. Thus S(n) is true for all positive integers n.3. S(4) : 4! = 24 > 16 = 2 4 is true. Assume S(k) : k! > 2 k is true. Then(k + 1)! = k!(k + 1) > 2 k · 2 = 2 k+1 , so S(k + 1) is true. Thus S(n) is truefor all positive integers n.8. Look at the proof in Example 3.11. S(0) : (1 + x) 0 − 1 = 0 ≥ 0 = 0 · x is true. Assume S(k) : (1 + x) k − 1 ≥ kx istrue. Then (1 + x) k+1 − 1 = (1 + x)(1 + x) k − 1 = (1 + x) k + x(1 + x) k − 1 ≥kx + x(1 + x) k ≥ kx + x = (k + 1)x, so S(k + 1) is true. Thus S(n) is truefor all positive integers n.15. (a) (14)14 + (−5)39 = 1.(c) (3709)1739 + (−650)9923 = 1.(e) (881)23771 + (−1050)19945 = 1.17. (b) Use mathematical induction. (c) Show that f 1 = 1, f 2 = 1, and f n+2 =f n+1 + f n . (d) Use part (c). (e) Use part (b) and Problem 16.19. Use the Fundamental Theorem of Arithmetic.23. Let S = {s ∈ N : a | s, b | s}. S ≠ ∅, since |ab| ∈ S. By the Principle ofWell-Ordering, S contains a least element m. To show uniqueness, supposethat a | n and b | n for some n ∈ N. By the division algorithm, there existunique integers q and r such that n = mq +r, where 0 ≤ r < m. a | m, b | m,a | n, b | n ⇒ a | r, b | r ⇒ r = 0 by the minimality of m. Therefore, m | n.27. Since gcd(a, b) = 1, there exist integers r and s such that ar + bs = 1 ⇒acr + bcs = c. Since a | a and a | bc, a | c.29. Let p = p 1 p 2 · · · p k + 1, where p 1 = 2, p 2 = 3, . . . , p k are the first k primes.Show that p is prime.