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Chapter 6
Chapter 6
Chapter 6
Chapter 6
Chapter 6
Chapter 6
Chapter 6
Chapter 6
Chapter 6
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Chapter 6

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Chapter 6

Chapter 6

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  • 1. Student ID: U10011024 Name: Kuan-Lun Wang75 Theorem. Eevery integer greater than 1 is a productof prime numbers. (P.34) We use proof by contradiction. Assume that some positiveinteger cannot be written as the product of primes. Let n bethe smallest such integer (such an integer must exist, fromthe well-ordering property). If n is prime, it is obviously theproduct of set a set of primes, namely the one prime n. So nmust be composite. Let n = ab, with 1<a<n and 1<b<n.But because a and b are smaller than n, they must be theproduct of primes. Then, because n = ab, we conclude thatn is also a product of primes. This contradiction shows thatevery positive integer can be written as the product of primes.Pratice 73. If p | a1a2 · · · an, then p | ai for some 1 ≤ i ≤n. (P.34)Lemma 3.5. If p divides a1a2 · · · an, where p is a prime anda1, a2, . . . , an are positive integers, then there is an integeri with 1 ≤ i ≤ n such that p divides ai. (P.113) We prove this result by induction. The case where n = 1is trivial. Assume that the results is true for n. Considera product of n + 1 integers a1a2 · · · an+1 that is divisible bythe integer p. We know that either (p, a1a2 · · · an) = 1 or(p, a1a2 · · · an) = p. If (p, a1a2 · · · an) = 1, then by Lemma3.4, p | an+1. On the other hand, if p | a1a2 · · · an, using theinduction hypothesis, there is an integer i where 1 ≤ i ≤ nsuch that p | ai. Consequently, p | ai for some i with 1 ≤i ≤ n + 1. This prove the result.CHAPTER 6 1
  • 2. Student ID: U10011024 Name: Kuan-Lun Wang Lemma 3.4. If a, b, and c are positive integers such that(a, b) = 1 and a | bc, then a | c. (P.34)76 Theorem (Fundamental Theorem of Arithmetic). Ev-ery integer greater than 1 can be represented as a product ofprimes in only one way, apart from the order of the factors. We arrange the prime factorization of n as follows: a n = p a1 p a2 · · · p k k , 1 2where a1>0, a2>0, . . . , ak >0 and p1<p2< · · · <pk . Thisfactorization of n is called the canonical factorization of n.For example 23 · 32 · 5 · 72 is the canonical factorization of17640. (P.34)Theorem 3.15. The fundamental Theorem of Arith-metic. Every positive integer greater then 1 can be writtenuniquely as product of primes, with the prime factors in theproduct written in nondecreasing order. (P.112) We use proof by contradiction. Assume that some positiveinteger cannot be written as the product of primes. Let n bethe smallest such integer (such an integer must exist, fromthe well-ordering property). If n is prime, it is obviously theproduct of set a set of primes, namely the one prime n. So nmust be composite. Let n = ab, with 1<a<n and 1<b<n.But because a and b are smaller than n, they must be theproduct of primes. Then, because n = ab, we conclude thatn is also a product of primes. This contradiction shows thatevery positive integer can be written as the product of primes. We now finish the proof of the fundamental theorem ofarithmetic by showing that the factorization is unique. Sup-CHAPTER 6 2
  • 3. Student ID: U10011024 Name: Kuan-Lun Wangpose that there is an integer n that has two different factor-izations into primes: n = p 1 p2 · · · ps = q 1 q 2 · · · q t ,where p1, p2, . . ., ps, and q1, q2, . . ., qt are all primes, withp1 ≤ p2 ≤ · · · ≤ ps and q1 ≤ q2 ≤ · · · ≤ qt. Remove all common primes from the two factorizations toobtain pi1 pi2 · · · piu = qj1 qj2 · · · qjv ,where the primes on the left-hand side of this equation differfrom those on the right-hand side, u ≥ 1, and v ≥ 1 (be-cause the two original factorizations were presumed to differ).However, this leads to a contradiction of Lemma 3.5; by thislemma, pi must divide qjk for some k, which is impossible,because each qjk is prime and is different from pi1. Hence,the prime factorization of a positive integer n is unique. Lemma 3.5. If p divides a1a2 · · · an, where p is a primeand a1, a2, . . . , an are positive integers, then there is aninteger i with 1 ≤ i ≤ n such that p divides ai. (P.113)78 Theorem. For a, b ∈ Z+, a · b = (a, b) · [a, b]. (P.35)Theorem 3.16. If a and b are positive integers, then[a, b] = ab/(a, b), where [a, b] and (a, b) are the least commonmultiple and greater common divisor of a and b, respectively.(P.116) Let a and b have prime-power factorizations a = pa1 pa2 · · · pan 1 2 n b1 b2 bnand b = p1 p2 · · · pn , where the exponents are nonnega-tive integers and all primes occurring in either factorizationCHAPTER 6 3
  • 4. Student ID: U10011024 Name: Kuan-Lun Wangoccur in both, perhaps with 0 exponents. Now let Mj =max(aj , bj ) and mj = min(aj , bj ). Then we have (a, b) · [a, b] = pM1 pM2 · · · pMn · pm1 pm2 · · · pmn 1 2 n 1 2 n = pM1+m1 p2 2+m2 · · · pMn+mn 1 M n = pa1+b1 p22+b2 · · · pan+bn 1 a n = p a1 p a2 · · · p an · p b 1 p b 2 · · · p b n 1 2 n 1 2 n = a · b,because Mj + mj = max(aj , bj ) + min(aj , bj ) = aj + bj byLemma 3.6. Lemma 3.6. If x and y are real numbers, then max(x, y)+min(x, y) = x + y. √79 Proposition. Prove that √ is irrational. (P.35)√ 2Example 3.20. Suppose that 2 is rational. Then 2 =a/b, where a and b are relatively prime integers with b = 0.It follows that 2 = a2/b2, so that 2b2 = a2. Because 2 | a2, itfollows (see Exercise 40 at the end of this section) that 2 | a.Let a = 2c, so that b2 = 2c2. Hence, 2 | b2, and by Exercise40, 2 also divides b. However, because (a, b) = 1, we knowthat 2 cannot divide both a and b. This contradiction shows √that 2 is irrational. (P.119) Exercise 40. Show that if a, b, and c are integers withc | ab, then c | (a, c)(b, c). (P.123)80 Theorem. Let α be a real root of the polynomial xn +cn−1xn−1 + · · · + c1x + c0, where the coefficients c0, c1, . . .,cn−1 are integers. Then α is either an integer or an irrationalCHAPTER 6 4
  • 5. Student ID: U10011024 Name: Kuan-Lun Wangnumber. (P.36)Theorem 3.18. Let α be a real number that is a root ofthe polynomial xn + cn−1xn−1 + · · · + c1x + c0, where thecoefficients c0, c1, . . ., cn−1 are integers. Then α is either aninteger or an irrational number. (P.119) Suppose that α is rational. Then we can write α = a/b,where a and b are relatively prime integers with b = 0. Be-cause α is a root of xn + cn−1xn−1 + · · · + c1x + c0, we have (a/b)n + cn−1(a/b)n−1 + · · · + c1(a/b) + c0 = 0.Multiplying by bn, we find that an + cn−1an−1b + · · · + c1abn−1 + c0bn = 0.Because an = b(−cn−1an−1 − · · · − c1abn−2 − c0bn−1),we see that b | an. Assume that b = ±1. Then b has a primedivisor p. Because p | b and b | an, we know that p | an.Hence, by Exercise 41, we see that p | a. However, because(a, b) = 1, this is a contradiction, which shows that b = ±1.Consequently, if α is rational then α = ±a, so that α mustbe an integer. Exercise 41. (P.123)a) Show that if a and b are positive integers with (a, b) = 1,than (an, bn) = 1 for all positive integers n.b) Use part (a) to proof that if a and b are integers such thatan | bn, where n is a positive integer, then a | b.CHAPTER 6 5
  • 6. Student ID: U10011024 Name: Kuan-Lun Wang n84 Definition. The integers Fn = 22 + 1 are called theFermat numbers. For example, F0 = 3, F1 = 5, F2 = 17, F3 = 257, andF4 = 65537.These numbers Fi, 0 ≤ i ≤ 4, are all primes. (P.37)The Fermat Number nThe integers Fn = 22 + 1 are called the Fermat numbers.Fermat conjectured that these integers are all primes. Indeed,the first few are primes, namely, F0 = 3, F1 = 5, F2 = 17, 5F3 = 257, and F4 = 65, 537. Unfortunately, F5 = 22 + 1 iscomposite, as we will now demonstrate. (P.131) 585 Proposition. The Fermat number F5 = 22 + 1 isdivisible by 641. (P.37) 5Example 3.24. The Fermat number F5 = 22 + 1 is di-visible by 641. We can show that 641 | F5 without actuallyperforming the division, using several not-so-obvious obser-vations. Note that 641 = 5 · 27 + 1 = 24 + 54.Hence, 522 + 1 = 232 + 1 = 24 · 228 + 1 = (641 − 54)228 + 1 641 · 228 − (5 · 27)4 + 1 = 641 · 228 − (641 − 1)4 + 1 641(228 − 6413 + 4 · 6412 − 6 · 641 + 4).Therefore, we see that 641 | F5. (P.131-132)86 Theorem. For all positive integers n, we have F0F1F2 · · · Fn−1 = Fn − 2.CHAPTER 6 6
  • 7. Student ID: U10011024 Name: Kuan-Lun Wang(P.37) kLemma 3.10. Let Fk = 22 + 1 denote the kth Fermatnumber, where k is a nonnegative integer. Then for all posi-tive integers n, we have F0F1F2 · · · Fn−1 = Fn − 2.(P.133-134) We will prove the lemma using mathematical induction.For n = 1, the identity reads F0 = F1 − 2.This is obviously true, because F0 = 3 and F1 = 5. Now, letus assume that the identity holds for the positive integer n,so that F0F1F2 · · · Fn−1 = Fn − 2.With this assumption, we can easily show that the identityholds for the integer n + 1, because F0F1F2 · · · Fn−1Fn = (F0F1F2 · · · Fn−1)Fn n n = (Fn − 2)Fn = (22 − 1)(22 + 1) n = (22 )2 − 1 = 22n+1 − 1 = Fn+1 − 2. This leads to the following theorem.Pratice 81. Prove that the least digit in the decimal ex- npansion of Fn = 22 + 1 is 7 if n ≥ 2. (P.37)Exercise 17. Show that the last digit in the decimal expan- nsion of Fn = 22 +1 is 7 if n ≥ 2. (Hint: Using mathematical ninduction, show that the last decimal digit of 22 is 6.) (P.136,622)CHAPTER 6 7
  • 8. Student ID: U10011024 Name: Kuan-Lun Wang We can prove that the last digit in the decimal expansion ofFn is 7 for n ≥ 2 by proving that the last digit in the decimal nexpansion of 22 is 6 for n ≥ 2. This can be done using 2mathematical induction. We have 22 = 16, so the result istrue for n = 2. Now assume that the last decimal digit of n n n+122 is 6, that is, 22 ≡ 6 (mod 10). It follows that 22 = n+1 −2n 2n 22 n+1 −2n(2 ) ≡ 62 ≡ 6 (mod 10). This completes theproof.87 Theorem. Let m and n be distinct nonnegative inte-gers. Then (Fm, Fn) = 1. (P.38)Theorem 3.21. Let m and n be distinct nonnegative in-tegers. Then the Fermat numbers Fm and Fn are relativelyprime. (P.134) Let us assume that m<n. By Lemma 3.10, we know that F0F1F2 · · · Fn−1 = Fn − 2.Assume that d is a common divisor of Fm and Fn. Then,Theorem 1.8 tells us that d | (Fn − F0F1F2 · · · Fm · · · Fn−1) = 2.Hence, either d = 1 or d = 2. However, because Fm and Fnare odd, d cannot be 2. k Lemma 3.10. Let Fk = 22 + 1 denote the kth Fer-mat number, where k is a nonnegative integer. Then for allpositive integers n, we have F0F1F2 · · · Fn−1 = Fn − 2.CHAPTER 6 8
  • 9. Student ID: U10011024 Name: Kuan-Lun Wang(P.133-134) Theorem 1.8. If a, b, and c are integers with a | b andb | c, then a | c. (P.37)CHAPTER 6 9

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