Discrete Mathematics with Applications MATH236 Dr. Hung P. - PowerPoint PPT Presentation

Discrete Mathematics with Applications MATH236 Dr. Hung P. Tong-Viet School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 Tong-Viet (UKZN) MATH236 Semester 1, 2013 1 / 8

Table of contents Mathematical Induction 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 2 / 8

Mathematical Induction Principle of Mathematical Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verity that P (1) is true Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 8

Mathematical Induction Principle of Mathematical Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verity that P (1) is true 2 Inductive Step: Show that if P ( k ) is true, then P ( k + 1) is true for all integers k ≥ 1 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 8

Mathematical Induction Principle of Mathematical Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verity that P (1) is true 2 Inductive Step: Show that if P ( k ) is true, then P ( k + 1) is true for all integers k ≥ 1 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 8

Mathematical Induction Example Example Use mathematical induction to show that for all nonnegative integer n , 1 + 2 + · · · + 2 n = 2 n +1 − 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 8

Mathematical Induction Example Proof. Let P ( n ) be the statement that 1 + 2 + · · · + 2 n = 2 n +1 − 1 for the integer n . Basis Step: P (0) is true because 2 0 = 2 1 − 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction Example Proof. Let P ( n ) be the statement that 1 + 2 + · · · + 2 n = 2 n +1 − 1 for the integer n . Basis Step: P (0) is true because 2 0 = 2 1 − 1 Inductive Step: Suppose that P ( k ) is true for an arbitrary nonnegative integer k . That is, we assume that 1 + 2 + · · · + 2 k = 2 k +1 − 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction Example Proof. Let P ( n ) be the statement that 1 + 2 + · · · + 2 n = 2 n +1 − 1 for the integer n . Basis Step: P (0) is true because 2 0 = 2 1 − 1 Inductive Step: Suppose that P ( k ) is true for an arbitrary nonnegative integer k . That is, we assume that 1 + 2 + · · · + 2 k = 2 k +1 − 1 We must show that P ( k + 1) is true, that is, we must show that 1 + 2 + · · · + 2 k +1 = 2 k +2 − 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction Example Proof. Let P ( n ) be the statement that 1 + 2 + · · · + 2 n = 2 n +1 − 1 for the integer n . Basis Step: P (0) is true because 2 0 = 2 1 − 1 Inductive Step: Suppose that P ( k ) is true for an arbitrary nonnegative integer k . That is, we assume that 1 + 2 + · · · + 2 k = 2 k +1 − 1 We must show that P ( k + 1) is true, that is, we must show that 1 + 2 + · · · + 2 k +1 = 2 k +2 − 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction Example We have that 1 + 2 + · · · + 2 k + 2 k +1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction Example We have that 1 + 2 + · · · + 2 k + 2 k +1 (1 + 2 + · · · + 2 k ) + 2 k +1 = Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction Example We have that 1 + 2 + · · · + 2 k + 2 k +1 (1 + 2 + · · · + 2 k ) + 2 k +1 = (2 k +1 − 1) + 2 k +1 = Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction Example We have that 1 + 2 + · · · + 2 k + 2 k +1 (1 + 2 + · · · + 2 k ) + 2 k +1 = (2 k +1 − 1) + 2 k +1 = 2 · 2 k +1 − 1 = 2 k +2 − 1 = By mathematical induction, we know that P ( n ) is true for all integers n ≥ 0 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction Example We have that 1 + 2 + · · · + 2 k + 2 k +1 (1 + 2 + · · · + 2 k ) + 2 k +1 = (2 k +1 − 1) + 2 k +1 = 2 · 2 k +1 − 1 = 2 k +2 − 1 = By mathematical induction, we know that P ( n ) is true for all integers n ≥ 0 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction More examples Example 1 Prove that 7 n +2 + 8 2 n +1 is divisible by 57 for every integers n ≥ 0 . 2 Prove that n ! > 2 n for all integers n ≥ 4 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 8

Mathematical Induction More examples Example 1 Prove that 7 n +2 + 8 2 n +1 is divisible by 57 for every integers n ≥ 0 . 2 Prove that n ! > 2 n for all integers n ≥ 4 3 Prove that 1 · 1! + 2 · 2! + · · · + n · n ! = ( n + 1)! − 1 for every integer n ≥ 1 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 8

Mathematical Induction More examples Example 1 Prove that 7 n +2 + 8 2 n +1 is divisible by 57 for every integers n ≥ 0 . 2 Prove that n ! > 2 n for all integers n ≥ 4 3 Prove that 1 · 1! + 2 · 2! + · · · + n · n ! = ( n + 1)! − 1 for every integer n ≥ 1 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 8

Mathematical Induction Strong Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verify that P (1) is true Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 8

Mathematical Induction Strong Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verify that P (1) is true 2 Inductive Step: For all positive integer k ≥ 1 , if P (1) , P (2) , · · · P ( k ) is true, then P ( k + 1) is true. Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 8

Mathematical Induction Strong Induction Theorem To prove that P ( n ) is true for all positive integers n , where P ( n ) is a statement involving n , we need to complete two steps: 1 Basis Step: Verify that P (1) is true 2 Inductive Step: For all positive integer k ≥ 1 , if P (1) , P (2) , · · · P ( k ) is true, then P ( k + 1) is true. Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 8