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

1

Mathematical Induction

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 + · · · + 2n = 2n+1 − 1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 8

Mathematical Induction

Example

Proof. Let P(n) be the statement that 1 + 2 + · · · + 2n = 2n+1 − 1 for the integer n. Basis Step: P(0) is true because 20 = 21 − 1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction

Example

Proof. Let P(n) be the statement that 1 + 2 + · · · + 2n = 2n+1 − 1 for the integer n. Basis Step: P(0) is true because 20 = 21 − 1 Inductive Step: Suppose that P(k) is true for an arbitrary nonnegative integer k. That is, we assume that 1 + 2 + · · · + 2k = 2k+1 − 1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction

Example

Proof. Let P(n) be the statement that 1 + 2 + · · · + 2n = 2n+1 − 1 for the integer n. Basis Step: P(0) is true because 20 = 21 − 1 Inductive Step: Suppose that P(k) is true for an arbitrary nonnegative integer k. That is, we assume that 1 + 2 + · · · + 2k = 2k+1 − 1 We must show that P(k + 1) is true, that is, we must show that 1 + 2 + · · · + 2k+1 = 2k+2 − 1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction

Example

Proof. Let P(n) be the statement that 1 + 2 + · · · + 2n = 2n+1 − 1 for the integer n. Basis Step: P(0) is true because 20 = 21 − 1 Inductive Step: Suppose that P(k) is true for an arbitrary nonnegative integer k. That is, we assume that 1 + 2 + · · · + 2k = 2k+1 − 1 We must show that P(k + 1) is true, that is, we must show that 1 + 2 + · · · + 2k+1 = 2k+2 − 1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 8

Mathematical Induction

Example

We have that 1 + 2 + · · · + 2k + 2k+1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction

Example

We have that 1 + 2 + · · · + 2k + 2k+1 = (1 + 2 + · · · + 2k) + 2k+1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction

Example

We have that 1 + 2 + · · · + 2k + 2k+1 = (1 + 2 + · · · + 2k) + 2k+1 = (2k+1 − 1) + 2k+1

Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 8

Mathematical Induction

Example

We have that 1 + 2 + · · · + 2k + 2k+1 = (1 + 2 + · · · + 2k) + 2k+1 = (2k+1 − 1) + 2k+1 = 2 · 2k+1 − 1 = 2k+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 + · · · + 2k + 2k+1 = (1 + 2 + · · · + 2k) + 2k+1 = (2k+1 − 1) + 2k+1 = 2 · 2k+1 − 1 = 2k+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 7n+2 + 82n+1 is divisible by 57 for every integers n ≥ 0. 2 Prove that n! > 2n for all integers n ≥ 4 Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 8

Mathematical Induction

More examples

Example

1 Prove that 7n+2 + 82n+1 is divisible by 57 for every integers n ≥ 0. 2 Prove that n! > 2n 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 7n+2 + 82n+1 is divisible by 57 for every integers n ≥ 0. 2 Prove that n! > 2n 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