An Elementary Proof of Bertrands Postulate Daniel W. Cranston - PowerPoint PPT Presentation

An Elementary Proof of Bertrand’s Postulate Daniel W. Cranston Virginia Commonwealth University Davidson Math Coffee March 25, 2011

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n .

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper.

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n .

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000.

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001.

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. � 2 n � 4 n / (2 n ) ≤ n

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. � 2 n � � 4 n / (2 n ) ≤ = f ( p ) n p ≤ 2 n

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. � 2 n � � � � � 4 n / (2 n ) ≤ = f ( p ) = f ( p ) f ( p ) f ( p ) . n p ≤ 2 n p ≤ 2 n 2 n n < p ≤ 2 n 3 < p ≤ n 3

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. 1 � 2 n � � � � � 4 n / (2 n ) ≤ = f ( p ) = f ( p ) f ( p ) f ( p ) . n p ≤ 2 n p ≤ 2 n 2 n n < p ≤ 2 n 3 < p ≤ n 3

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. 1 1 � 2 n � � � � � 4 n / (2 n ) ≤ = f ( p ) = f ( p ) f ( p ) f ( p ) . n p ≤ 2 n p ≤ 2 n 2 n n < p ≤ 2 n 3 < p ≤ n 3

Overview Bertrand’s Postulate For every positive n , there exists a prime p s.t. n < p ≤ 2 n . Proof Outline � 2 n � Find upper and lower bounds for = (2 n )! / ( n ! n !). If no such p n exists, then lower bound is larger than upper. Only works for big n . Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n , let f ( p ) = p k s.t. p k | � 2 n and p k +1 � 2 n � � � | . n n If Bertrand’s is false, then ∃ n > 4000 s.t. 1 1 � 2 n � � � � � 4 n / (2 n ) ≤ = f ( p ) = f ( p ) f ( p ) f ( p ) . n p ≤ 2 n p ≤ 2 n 2 n n < p ≤ 2 n 3 < p ≤ n 3 So, we need bounds on f ( p ). . .

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k Ex: p = 3, n = 17.

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 � � � � 2 n n Cor: f ( p ) = p r , where r = � − 2 . k ≥ 0 p k p k

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 � � � � 2 n n Cor: f ( p ) = p r , where r = � − 2 . k ≥ 0 p k p k � � � � 2 n < 2 n n p k − 2( n 1. − 2 p k − 1) = 2. p k p k

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 � � � � 2 n n Cor: f ( p ) = p r , where r = � − 2 . k ≥ 0 p k p k � � � � 2 n < 2 n n p k − 2( n 1. − 2 p k − 1) = 2. p k p k 2. If p k > 2 n , then summand is 0, so f ( p ) ≤ 2 n .

Upper Bounds on f ( p ) � � n Legendre’s Thm: n ! contains factor p exactly � times. k ≥ 1 p k � 17 � 17 � 17 � � � Ex: p = 3, n = 17. So: + + + . . . = 5 + 1 + 0 = 6 . 3 2 3 3 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 � � � � 2 n n Cor: f ( p ) = p r , where r = � − 2 . k ≥ 0 p k p k � � � � 2 n < 2 n n p k − 2( n 1. − 2 p k − 1) = 2. p k p k 2. If p k > 2 n , then summand is 0, so f ( p ) ≤ 2 n . √ 2 n , then p 2 > 2 n , so f ( p ) ≤ p . 3. If p >