Chapter 1 Basic Number Concepts

Transcription

1 Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11: a. 363 = Not divisible by 9. b. 4,576. Digit sum is 22, then 4, so not divisible by 3 or 9. Alternate digit sum is 0, so divisible by 11. 4,576 = c. 123,456,789. Digit sum is 45, so divisible by 3 and 9. Alternate digit sum is 5, so not divisible by ,456,789 = 13,717, Logic and Proofs 2. The equivalent groupings are a and e ( A B ), b and h ( A B ), c and g ( B A ), and d, f and i ( B A ). Note that these are the four possible implication relationships between two statements A and B. 3. a = 1 a a a. 4. a b m Z s.t. b = a m b = a ( m) a ( b). 5. a b m Z s.t. b = a m. a, b positive m positive m 1 0 (m 1) a 0 b a 0 b a. 6. a b m Z s.t. b = a m. a c n Z s.t. c = a n. Thus, b+c = am + an = a (m+n) a (b+c). 7. a b n Z s.t. b = a n b m = (a n) m = a (n m) for any integer m a bm. 8. Combine the proofs for Problems 4, 6 and This is false. For example, 4 36 = 2 18, but 4 is not a factor of either 2 or 18. It is true if the divisor is prime, as we will prove in Chapter n odd n = 2m +1 n 2 = 4m 2 + 4m +1= 4[m(m +1)] +1. Either m or m+1 is even, so 4m(m+1) is a multiple of 8, which implies 8 (n 2 1). 11. The proof is essentially the same as that for divisibility by 3 or 9. Again, we give the proof initially for a 4-digit number. Let N = 1000d + 100c + 10b + a. The alternating sum of the digits, which we can call S, is d + c b + a. The difference between N and S, which we can call D, is 1001d + 99c + 11b. Thus, N = S + D. (We could have started the alternating sum with a positive sign, in which case we would have added N and S to get the same value for D.) As in the proof for divisibility by 3, D is always a multiple of 11, no matter what the values of d, c, b and a,

2 because 1001, 99 and 11 are all divisible by 11. Hence, as in the proof for divisibility by 3, N is a multiple of 11 iff D is a multiple of 11. It is a bit harder than in the case of 3 to see that this would remain true no matter how many digits were in N. This depends on every even power of 10 being 1 more than a multiple of 11 and every odd power of 10 being 1 less than a multiple of 11. This is easy using modular arithmetic, which is presented in Chapter 5. One need only observe that 10 1 (mod 11). To prove this using algebra, observe that 1 is always a root of the equation x n 1 = 0 for n even and is always a root of the equation x n + 1 = 0 for n odd. So (x+1) is a factor of x n 1 for n even and of x n + 1 for n odd. Now substitute 10 for x. S1-2

3 1.4. Division and Remainders 12. Quotient = 13,698. Remainder = There are many such sets. The set of all positive rational numbers has no smallest element. Other examples include the set of all of the reciprocals of the positive integers, or any sequence of the form a, a 2, a 3, a 4,, where a is a rational number less than The proposition is obviously true for n = 1: 1 = 1 2. Now suppose that n 1 = n 2. Add 2n+1 to both sides. The left side becomes the sum of the first n+1 odd numbers. The right side becomes (n+1) 2. This completes the proof by induction. 15. Assume that a proposition about positive integers P(n) is true for n = 1 and that if P(n) is true then P(n+1) is true. Let S be the set of positive integers for which the proposition P(s) is not true. If S is not empty, it must have a smallest element. Call that number s. Now s cannot be 1 because P(1) is true. Therefore s 1 is a positive integer and because s is the smallest element of S, s 1 is not in S. Therefore P(s 1) is true. But since the truth of P(n) implies the truth of P(n+1) for any positive integer n, the truth of P(s 1) implies the truth of P(s). But this contradicts the fact that s is in S. Therefore S is empty Greatest Common Factor and Least Common Multiple 16. The common factors of 30 and 70 are 1, 2, 5 and 10. The set of all common factors is the same as the set of factors of the greatest common factor, which is The common factors of 70 and 315 are 1, 5, 7 and 35. Once again, the set of all common factors is the same as the set of factors of the greatest common factor, which is = = If g is the gcf of a and b, the numbers a g and b g are relatively prime because if d were a common factor of a g and b g, dg would be a common factor of a and b. This would contradict the fact that g is the gcf of a and b. S1-3

4 20. Suppose m is the LCM of two integers a and b and M is another common multiple of a and b. If m does not divide M, then we can divide M by m to obtain a quotient q and a positive remainder r that is less than m. Now r = M mq, and M and mq are both common multiples of a and b, so their difference r is also a common multiple of a and b, which contradicts that m is the least common multiple. Therefore, m M. 21. The LCM of two relatively prime integers a and b is simply their product ab. Proof: The product clearly is a common multiple. If m is the LCM of a and b, then from the preceding problem, m ab, so ab = mx. Since m is a multiple of a, we can write m = as, and similarly m = bt. Therefore, ab = mx = asx, which implies that b = sx. Similarly, a = tx. Together, these equations show that x is a common divisor of a and b. Since (a,b) = 1, this means that x = 1 and therefore ab = m. 22. Suppose that the LCM of x and y is m. Then km is clearly a common multiple of kx and ky. Now suppose M is another common multiple of kx and ky. Then there exist integers a and b s.t. M = kxa and M = kyb. From this it follows that ax = by, so this number is a common multiple of x and y. Since m is the LCM of x and y, it follows that m is a factor of ax = by, which means that km is a factor of kax = kby = M. Thus, km divides any common multiple of kx and ky, which completes the proof that km is the LCM of kx and ky. 23. The relationship of the LCM of any two integers to their gcf is LCM[a,b] = ab a. From Problems 19 and 21, we know that (a,b) (a,b) and b ab are relatively prime so their LCM is the product. In the 2 (a,b) (a,b) preceding problem we showed that if the LCM of x and y is m, then the LCM of kx and ky is km. So, setting k = (a,b), x = a (a,b) and y = b (a,b), we have shown that the LCM of a and b is ab (a,b). S1-4

5 1.6. Prime Numbers = = is prime , so you need to check for prime factors up to 140. The reward comes near the end: 19,177 = The square of a prime is the smallest composite number for which that prime is the smallest prime factor. 27. The last composite number crossed off is 169 = The next prime after 13 is 17 and its square exceeds There are 25 primes between 1 and 100. There are 21 primes between 101 and There are 14 primes between 900 and There are 15 primes between 4900 and However, there are only 8 primes between 4800 and This shows that the decline in the density of primes is irregular. 30. The largest gap between primes less than 200 is from 113 to There are strings of consecutive composite numbers of any length. See Problem The number of primes is infinite. 33. Suppose the set of integers greater than 1 that have no prime divisors is not the empty set. Then by the well ordering principle, that set has a smallest element. Say that number is n. If n is prime, then it has itself as a prime divisor, since every integer divides itself, so n cannot be prime. Therefore n can be factored into two smaller integers a and b: n = ab. Because a and b are each smaller than n, each has a prime factor (n was the smallest integer without a prime factor). But any prime factor of a or b is also a prime factor of n, which contradicts that n has no prime factor. So the set of integers with no prime factor is empty. 34. Suppose the set of integers greater than 1 that have no prime factorization is not the empty set. Then by the well ordering principle, that set has a smallest element. Say that number is n. If n is prime, then its prime factorization is simply the number n, so n cannot be prime. Therefore n can be factored into two smaller integers a and b, neither of which is equal to 1: n = ab. Because a and b are each smaller than n, each has a prime factorization (n was the smallest integer without a prime factorization). So the product of the prime factorizations of a and b constitutes a prime factorization of n, which contradicts that n has no prime factorization. So the set of positive integers greater than 1 with no prime factorization is empty. S1-5

6 35. If m is positive and composite then there are positive integers a and b between 1 and m such that m = ab. Either a or b must be less than or equal to m, for otherwise their product would exceed m. Say a m. From Problem 34, a has a prime factor and from Problem 5, that factor cannot exceed a. That prime factor of a is also a prime factor of m, which completes the proof = 30,031 = The value of n 2 n + 41 is indeed prime for the first 40 positive integer values of n, which is rather remarkable, but the value of the expression for n = 41 is 41 2 = 1681, which obviously is not prime. 38. In order for a polynomial f(n) to generate a prime value for n = 0, its constant term must be a prime; say the constant term is p. If one sets n = kp for different integer values of k, the value of f(kp) must always be divisible by p, since every term of f(kp) is divisible by p. If f(kp) is always prime, then f(kp) must be equal to p for every value of k. But this is impossible, since a polynomial of degree m can take on a given value at most m times (for otherwise the polynomial f(n) p would have more than m roots). Therefore, no polynomial f(n) can generate a prime for every value of n. 39. n!+2 is divisible by 2, n!+3 is divisible by 3, n!+4 is divisible by 4, n!+n is divisible by n. 40. lim x ( ) li x x /ln(x) = lim x D( li( x) ) ( ) = lim 1 ln(x) x ( ln(x) 1) ( ln(x) ) = lim 2 x D x /ln(x) 1.7. Basic Properties of Integers No problems in this section. ln(x) ln(x) 1 =1. S1-6