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Unformatted text preview: 11/ 16/07 Suppose S is a set of a positive integer n0. Also suppose: 1. n S 2. If for each integer m, with n m n +2, m S, then n +2 S. Then S = {k Z+  k n } n 0 S n +1 S n +2 S n +3 S . . . n 0+1028023 S Strong principle Mathematic Induction Its logically equivalent to the ordinary PMI & the generalized PMI. A positive p>= 2 is prime means if it p=ab, where a & b are integers, then one of a & b is 1 and the other is p. A prime factorization of a positive integer n2 consists of: 1. positive integer k & 2. Prime numbers P 1 ,P 2 ,.,P k such that: n = P 1 ,P 2 ,,P k 12 = 2*2*3 k = 3 P 1 =2 P 3 =3 17= 17 k = 1 P 1 =17 15= 3*5 k = 2 P 1 =3, P 2 =5 Theorem Suppose n is a positive integer 2. Then n has a prime factorization. Proof using strong Induction. Base case is n=2 : If n=2, we can choose k=1, P1=2, since 2 is prime. So if n=2, n has a prime factorization. Induction step : Suppose n is positive integer 2 such that every integer m satisfy 2<=m<=n has a prime factorization. We want to prove that n+1 has a prime factorization. We consider 2 cases: n+1 prime & n+1 composite. Case 1 : If n+1 is prime, then we can choose k=+1, p 1 =n+1 and n+1 =p 1 , is the desired prime factorization. Case 2 : n+1 is not prime. In this case, there exist positive integer a & b such that 2 a,b n and n +1 = ab So by the IH, both a & b have prime factorizations. So there exist positive integers u & v & primes q 1 ,q 2 ,..,q u & r 1 ,r 2 ,,r u such that P 1 ,P 2 ,,P n a = q 1 ,q 2 ,.q n and b = r 1 ,r 2 ,..,r u So we can choose k= u+v and define: Pi = q i , for 1 i n P i= r iu , for u+1 i u+v P u+1 = r n+1u = r 1 P u+2 = r u+2u = r 2 . . P k = P n+v = ru+vu = rv Fibonacci numbers f (n) = 1, if n=1,2 f n1 + f n2 , if n 3 n f n 1 1 2 1 3 f 3 = f 2 +f 1 = 1+1 = 2 4 f 4 = f 3 +f 2 = 2+1 = 3 5 f 5 = f 4 +f 3 = 3+1 = 5 6 f 6 = f 5 +f 4 = 5+3 = 8 7 f 7 = f 6 +f 5 = 8+5 =13 What is the growth rate of the Fibonacci numbers? (n 2 ) ? (n) ? (b n ) for some real number b>1. So they grow faster than an exponential function. f n b n+k for some real number b and some integer k b = 1+ 5 / 2 f n is (b n ) Find positive constants k & n such that f n k b n for all n> n Proof of conjecture Base case : We need to see that f 3 b 32 = b but 11/21/07 Homework Solution: Suppose n is an integer > 6. Then 3 n < 1 Base case: If n=7, 3 7 = 2187. Also if n=7, n! = 5040....
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 Fall '07
 PeterPacheco
 Math

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