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Answers and Hints to the Exercises Answers and Hints to the Exercises Chapter 1 2 a) transitivity fails; b) the three equivalence classes are 3Z,1 + 3Z,2 + 3Z;c) reflexivity fails for 0; d) transitivity fails; e) symmetry fails. Chapter 2 2. Add (k + 1)3 to both sides of P(k), then show that k(k + 1) (k + 1)(k + 2) ( )2 +(k + q)3 =( )2. 2 2 5. Write + + xk 1 − 1 =(xk 1 − x)+(x − 1)=x(xk − 1)+(x − 1) and substitute for xk − 1usingP(k) 8. (k + 1)! =(k + 1)k! > (k + 1)2k > 2 · 2k = 2k+1. 9. First show that for k ≥ 4, (k + 1)4 = k4 + 4k3 + 6k2 +(4k + 1) ≤ 4k4. 10.Forc)usethat 1 = 1 < 1 . 2 (2n + 1) 8tn + 1 8tn 11. Use Exercise 5 with x = a. 13. 16n − 16 = 16(16n−1 − 1): can use Exercise 5 with x = 16. 14. Can use Exercise 5 with x = 8/3. 15. Can use Exercise 5 with x = 34. 569 570 Answers and Hints to the Exercises 16. Write 4k+2 = 4k(52 − 9). 17. Write 22k+3 = 22k+1 · (3 + 1). 20. First observe a1 = a (why?). Then for each m, prove P(n) : am+n = aman for n ≥ 1 by induction on n. 21. If N(n) is the number of moves needed to move n disks from one pole to another, show N(n + 1)=N(n)+1 + N(n). 22. For 4 disks, the answer is 80. 23. If r2 < n ≤ (r + 1)2,then(r + 1)2 < 2n? 24. Check the argument for n = 1. 26. See the proof of Theorem 6. 27. Adapt the proof of Proposition 4. 28. Divide the polygon into two polygons by an edge connecting two non-adjacent vertices. 29. Let P(n) be “For some k ≥ 1, f (k)(n)=1,” and prove that P(n) is true for all n ≥ 1. 30. See Section 3A, Example 1. + 5 < ( 5 )2 31. Show that 1 3 3 . 32. Try observing that c(n + 1)=1 + c(1)+c(2)+...+ c(n). 33. Let S be the set of numbers a > 0 for which there is a number b with 2a2 = b2. 35. Given a non-empty descending chain of natural numbers, let S be the set of numbers in the chain. 37. Set x = y = 1 in the Binomial Theorem. 38. Fix s and do it by induction on n ≥ s, using Corollary 13. 39. Write (x+y)2n =(x+y)n(x+y)n, expand (x+y)n and (x+y)2n by the Binomial Theorem and collect the coefficients of xnyn. Chapter 3 1. a) If a is a least element of S,thena ≤ s for all s in S;ifb is a least element of S, then b ≤ s for all s in S,sob ≤ a. But since b is in S, a ≤ b. b) ris the least element of S = {b − aq | q in N}.Sor is unique. = b r = b 2. q a , a a . 3. a = ds,b = dt implies r = b − aq = dt − qds = d(t − qs). 4. Use Exercise 3. 5. Let d = ay + bz.Divided into a:ifa = dq+t,thena =(ay + bz)q +t,sot is in J.Ift > 0, then d is not the least element of J. Then repeat, dividing d into b. 7. (11111000011)2 =(3,7,0,3)8 . 10. 7855 seconds = 2 hrs, 10 mins, 55 seconds. 11. This problem involves distance in base 1760 and time in base 60. 15. With n = 4, adding the five products of “digits” <104 can exceed 108.So10n with n = 3 is the largest possible base if you do both addition and multiplication on the calculator. Answers and Hints to the Exercises 571 19. a): 5; c): 1 . 20. 17, but see Exercise 33. 22. See Exercise 35. 23. Use (or redo) Exercise 3. 25. If d divides a and b,thend divides ar + bs, so.... (See also Corollary 6 of Section D.) 30. Let the m consecutive integers be a,a + 1,...,a +(m − 1).Ifa ≥ 0, then a = mq+r,andifr = 0, then m divides q.Ifr > 0thena+(m−r) ≤ a+(m−1) and is divisible by m.Ifa < 0 write −a = mq+s with 0 ≤ s < m.Ifs = 0thenm divides a. If s > 0, then a + s < b and a + s is divisible by m. (See also Section 6D.) 31. See Section 5B, Proposition 4. 32. Let m − n = d.Thenn,n + 1,...,n + d are d consecutive integers–apply Exercise 30. 35. e) 1. 37. Show that every common divisor of a and b is a divisor of r, hence a common divisor of a and r. Then show that every common divisor of a and r is a common divisor of a and b. 38. Try induction on the number of divisions in Euclid’s Algorithm. 40. One example is: 6 divides 2 · 3. 41. Generalize the examples in Exercise 40. 44. If ar + ms = 1andbt + mw = 1, then 1 =(ar + ms)(bt + mw)=aby + mz for y = rt and z = .... 46. Factor d from ra + sb and use Corollary 6. 48. Let d =(a,b).Thenmd divides ma and mb,somd ≤ (ma,mb). Also, let d = ar + bs,thenmd = mar + mbs,so(ma,mb) divides md. 49. If e =(ab,m),thene divides m and (m,b)=1, so (e,b)=1. Since e divides ab, Corollary 7 yields that e divides a. Thus e ≤ (a,m).Also,(a,m) ≤ (ab,m)=e. 51. Use Corollary 6. 52. Can assume r,s are integers with (r,s)=1. Try to show that s = 1. 57. b) x = 19 + 45k,y = −8 − 19k. 59. a) x = 13 + 31k,y = 7 + 17k with k ≥ 0 b) x = 18 + 31k,y = 10 + 17k with k ≥ 0. 61. d = 1, r = 731 + 1894k,s = −1440 − 3731k for all k in Z. 63. Find integer solutions of 5 f = 9c + 160. 65. Observe that 5 = 16 · 6 − 13 · 7. 69. Let b = aq1 + r1,a = r1q2 + r2 with r1 < a and q2 ≥ 1, and assume a < Fn.If r1 < Fn−1, then by induction, N(r,a) ≤ n − 4. If r1 ≥ Fn−1,then Fn−1 + Fn−2 = Fn > a ≥ r1 + r2 ≥ Fn−1 + r2, so r2 < Fn−2, hence by induction, N(r2,r1) ≤ n − 5. In either case, N(a,b) ≤ n − 3. 70. a) Do n = 0,1, then check that if the formula is true for n = k − 2andfor n = k − 1, then it is true for n = k. 572 Answers and Hints to the Exercises Chapter 4 2. For the induction step, let c = a1a2 · ...· an−1 and d = an and apply Lemma 3 to cd. 4. Since (n,q) ≤ (n,qr) in general, it suffices to show that if (n,q)=1then (n,qr)=1, or equivalently, if (n,qr) > 1then(n,q) > 1. If (n,qr) > 1, let p be a prime divisor of (n,qr).Thenp divides qr,sobyExercise2,p divides q.Sincep also divides n, p ≤ (n,q) so (n,q) = 1. √ 5. Show that n cannot factor into n = ab where both a and b are > n 6. Use Exercise 5 and note that 2021 < 452. 8. a) m in 3N is irreducible iff n = 3q with q not a multiple of 3. c) 30 · 3 =√6 · 15. √ √ + − = + − = − = a−c 9a)Ifa b 23 c d 23 with b d,then 23 d−b would be a ratio- nal number. But then the negative number −23 would be the square of a rational number, impossible.√ d) If a + b −23 divides 3, then there are integers c and d so that √ √ (a + b −23)(c + d −23)=3. This is true iff ac + 23bd = 3andad + bc = 0byparta),andthisinturnistrueiff √ √ (a − b −23)(c − d −23)=3. Multiplying the two equations together yields (a2 + 23b2)(c2 + 23d2)=9. The only solutions of this equation must have b = d = 0. 13. If 1001/5 = a/b,thenb5 · 22 · 52 = a5.If2e a and 2 f b,then5e + 2 = 5 f , impossible. 14 b) “integers” can be negative. 15. First show that for each prime p dividing c,ifpe c then per cr.Ifpg a,thenp does not divide b,sog = er.Soifa > 0, then a is an r-th power. Similarly for b. r r Finally, if a = −a0 where a0 > 0anda0 = d ,thena =(−d) because r is odd. 17. Be sure to check your answer! 19. Write a = 2r f ,b = 2sg where f ,g are odd and coprime. Then check the various possibilities for min{4r,5s} given that min{r,s} = 3. 21. Use Lemma 3 or Chapter 3, Corollary 6. 27. Write ar + bs = c and review the proof of Chapter 3, Corollary 8. 28. Let q be the product of all primes p dividing c such that (p,a)=1. Show that (a + bq,c)=1. 35. See Section 3E. 37. See Exercise 17. 39. Observe that if a divides bk then bk is a common multiple of a and b. 42. Look for examples where [a,b,c] < abc/(a,b,c). Answers and Hints to the Exercises 573 43. Note that if p1 = 4k1 + 1, p2 = 4k2 + 1, then p1 p2 = 4l + 1forsomel. 44. Every odd prime has the form p = 6k + 1orp = 6k − 1. 50. Show that every prime that divides n! + 1mustbe>n. Chapter 5 1. For a > 0 use Proposition 1. For a < 0 show there is a smallest k > 0sothat r = a + km ≥ 0, then show that 0 ≤ r < m.
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