Summary: Divisibility and Factorization

Transcription

1 Summary: Divisibility and Factorization One of the main subjects considered in this chapter is divisibility of integers, and in particular the definition of the greatest common divisor Recall that we have seen two ways to compute the greatest common divisor of two integers a and b: Comparing the list of divisors for a and the list of divisors for b Comparing the prime factorizations of a and b The most common way to compute the list of divisors for an integer n is to find first the prime factorization of n, and then use this factorization as a guide for computing the divisors (The procedure for doing this is described later in this summary) Thus, both methods above will require the prime factorization of a and b However, the second method bypasses the step of computing the lists of divisors required to apply the first method, and so the second method is typically more efficient than the first Your numerical investigations into Research Question 1 should have led to the conjecture given below Our practice here and in later chapters will be to state the conjecture, and then follow the statement with a proof 1 Theorem 13 (RQ1) The set of positive divisors of gcd(a, b) is equal to the intersection of the set of positive divisors of a and the set of positive divisors of b Proof We begin by noting that the proof requires us to show that two sets are equal We will do this in the standard manner, by showing that each set is a subset of the other, and then concluding that the sets must be equal (This technique is discussed in detail in the Tips for Writing Proofs section of the Course Guide) First, suppose that d = gcd(a, b), and that c is in the set of positive divisors of gcd(a, b) In other words, c>0 and c is a divisor of d Since d is a divisor of a and d is a divisor of b, it follows that c is a divisor of a and c is a divisor of b Thus c is in the set of divisors of a and c is in the set of divisors of b (and c is positive) Therefore c is in the intersection of the set of positive divisors of a and the set of positive divisors of b, and so we see that the set of positive divisors of gcd(a, b) isa subset of the intersection of the set of positive divisors of a and the set of positive divisors of b We now assume that c is in the intersection of the set of positive divisors of a and the set of positive divisors of b Then c is a divisor of a, c is a divisor of b, and c>0 Either c = 1, in which case we are done (because c = 1 is a divisor of d = gcd(a, b)), or c has a prime-power factorization Now suppose that the prime factorization of c is given by c = p e 1 1 p e 2 2 pe 1 We use the label Theorem because the statement of the conjecture is accompanied by a proof As discussed in the Course Guide, once a conjecture is proved, it becomes a theorem 85

2 86 CHAPTER 1 Then, by the Fundamental Theorem of Arithmetic, it follows that p e 1 1 must be a divisor of both a and b This in turn implies that p e 1 1 must be a divisor of d (This assertion can be seen to be true by looing at the formula for computing the greatest common divisor given in the GCDs and Factorization section of the electronic noteboo) The same argument as above suffices to show that p e 2 2,, p e are also divisors of d, and so it follows that c is a divisor of d Since c>0, c is in the set of positive divisors of gcd(a, b), and so the intersection of the set of positive divisors of a and the set of positive divisors of b is a subset of the set of positive divisors of gcd(a, b) Combining the results contained in the two paragraphs above shows that the set of positive divisors of gcd(a, b) is equal to the intersection of the set of positive divisors of a and the set of positive divisors of b The result of Research Question 1 is often used in the following form: Corollary 14 If a and b are integers which are not both 0, and c is an integer such that c j a and c j b, then c j gcd(a, b) This is a corollary because it is easy to prove using Research Question 1 Proof Either c will be a positive common divisor of a and b By RQ1, c is a positive divisor of gcd(a, b) Hence, c j gcd(a, b) Research Questions 2 to 6 all build up to the following general result: Theorem 15 (RQs2{6) Suppose that n = p a 1 2 pa Then d(n) =(a 1 + 1)(a 2 +1) (a +1) Proof If m is a positive divisor of n, then m has a prime factorization of the form m = p b 1 2 p b, where 0 b 1 a 1,0 b 2 a 2,,0 b a Therefore, we have (a 1 +1) possible choices for b 1,(a 2 + 1) possible choices for b 2, and so on Thus, the total possible number of choices is equal to the product of the choices for b 1, b 2,, b, which is (a 1 + 1)(a 2 +1) (a +1) Now, is it possible that two different choices for b 1, b 2,, b will result in the same value of m? If this did occur, then this value of m would have two different prime factorizations, a violation of the Fundamental Theorem of Arithmetic So, since each different choice for b 1, b 2,, b results in a distinct value of m, the result follows

3 87 Solutions to Selected Exercises The first exercise comes from the electronic noteboo introducing Mathematica or Maple Exercise 01 The following calculations suggest a pattern What is the general formula suggested by these calculations? Solution The general formula is (1 x)(1+ x) =1 x 2 (1 x)(1+ x + x 2 )=1 x 3 (1 x)(1+ x + x 2 + x 3 )=1 x 4 (1 x)(1+ x + x 2 + x 3 + x 4 )=1 x 5 (1 x)(1+ x + + x n )=1 x n+1 Although the exercise did not request a proof, we will supply one here One way of looing at this formula is to find a simple formula for the sum 1+ x + + x n It is a finite geometric series with common ratio x One could apply the formula for the sum of a geometric series The proof we setch here uses the same approach as the standard method for summing a geometric series The plan is to multiply 1+ x + + x n by x which gives x + x 2 + x n+1 If we subtract this from the original sum, most of the terms cancel Here is the computation: (1 x)(1+ x + + x n )=1+x + + x n x(1+ x + + x n ) =1+x + + x n (x + x 2 + x n+1 ) =1+x + + x n x x 2 x n+1 =1+x x + x 2 x x n x n x n+1 =1 x n+1 Next we loo at one of the exercises from the current chapter Exercise 12 Suppose n = p m 1 1 p m 2 terms of their prime power factorizations? 2 p m What are the positive divisors of n in Solution In the course of the proof of Research Questions 2 to 6, we have implicitly answered this question Specifically, if n = p m 1 1 p m 2 2 p m, then we just set d = p b 1 2 p b,

4 88 CHAPTER 1 and tae all possible combinations of b 1, b 2,, b, subject to the conditions that 0 b 1 m 1,0b 2 m 2,,0b m For example, suppose that we have n =40=2 3 5 From the precedingtheorem, we now that the number of positive divisors of n is equal to (3 + 1)(1 + 1) = 8 Moreover, the values of these divisors are = = = = = = = =40 We now give a formal proof of our statement Proof First suppose that d = p b 1 2 p b with 0 b i m i for each i from 1 to Then, if we let c = p m 1 b 1 1 p m 2 b 2 2 p m b, then each exponent for c is nonnegative Therefore, c 2 Z and cd = n Thus, d j n We now suppose that d is a positive divisor of n Thus there exists an integer c such that dc = n If d is divisible by a prime q where q is not one of the primes p i which occur in the factorization of n, then there is an integer d such that d q = d Substituting, we find that n = dc = d qc = qd c Writing d c as a product of primes, we would have expressed n as a product of primes which include q This would contradict the unique factorization of n Therefore, d cannot be divisible by any prime except those which occur in the factorization of n In particular, the prime factorization of d has the form d = p b 1 2 p b We currently have shown that the b i 0 We still need to prove that b i m i for all i From dc = n, we now that c is also a divisor of n, and it is positive since d and n are positive So, the preceedingdiscussion applies to c as well Thus we can write c = p e 1 1 p e 2 2 pe with 0 e i for all i Now n = dc = p b 1 2 p b p e 1 1 p e 2 2 pe = p b 1+e 1 +e 2 2 p b +e By the unique factorization of n, we have p b i+e i i = p m i i for all i Thus, b i + e i = m i Since e i 0 for all i, we have that b i m i for all i Going Farther: Sums of Divisors Now that the question of how to determine the number of positive divisors of an integer has been settled, we turn to a related question: what is the formula for the sum of the positive divisors of an integer? Suppose that n is a positive integer Then σ(n) denotes the sum of all positive divisors of n; that is, X σ(n) = d, d n

5 P where the notation d n indicates that the sum extends over all positive divisors d of n For example, we have X σ(40) = d = d 40 = 90 Our goal in what follows is to develop a formula for σ(n) that is analogous to the formula for d(n) To begin, suppose that n = p a, where p is a prime In this case it is easy to see that the positive divisors of n are 1, p, p 2,, p a, so that σ (p a )=1+p + p p a The sum on the right is geometric, and so can be condensed to a closed form by using the formula from exercise 01 above: 1+x + x x n = 1 xn+1 1 x = xn+1 1 x 1 (In this form, the formula requires x 6= 1) Thus, it follows that σ (p a )= pa+1 1 p 1 Next suppose that n = p a 1 2 Then σ(n) is equal to the sum of the divisors of n, which are listed below: p 0 1p 0 2, p 0 1p 1 2, p 0 1p 2 2, p 0 1p a 2 2, p 1 1p 0 2, p 1 1p 1 2, p 1 1p 2 2, p 1 1p a 2 2, p 2 1p 0 2, p 2 1p 1 2, p 2 1p 2 2, p 2 1p a 2 2, p a 1 1 p 0 2, p a 1 1 p 1 2, p a 1 1 p 2 2, p a 1 Now consider the product (1 + p 1 + p p a 1 1 )(1 + p 2 + p p a 2 2 ) If this product was multiplied out, then each term in the expanded sum would correspond with exactly one of the terms in the preceding list 2 Thus, it follows that σ (p a 1 2 ) = (1 + p 1 + p p a 1 1 )(1 + p 2 + p p a 2 2 ), 2 If you have trouble seeing that this is true, try an example For instance, set n = 40, write out the resulting table and product, and then loo for the correspondence 2 89

6 90 CHAPTER 1 and so we may conclude that σ (p a 1 2 )= p a p 1 1 p a p 2 1 With the previous special cases established, you can probably guess the pattern for the general case If n = p a 1 2 pa, then p a p a p a +1 1 σ(n) = (11) p 1 1 p 2 1 p 1 Aproof of this formula (which is included as a homewor exercise) can be constructed using induction and the same reasoning that was applied in the special cases