SUBTOPIC 5MATHEMATICAL

INDUCTION

UNIVERSITI PENDIDKAN SULTAN IDRIS

PREPARED BY : MOHAMAD AL FAIZ BIN SELAMAT

Principle of Induction. To prove that P (n) is true for all positive integers n, where P (n) is a propositional function. A proof by mathematical induction has two parts:

Basic step: We verify that P (1) is true. Inductive step: We show that the conditional

proposition ∀ k ∈ , P (k) ⇒ P (k + 1) is true.

Introduction

How would you prove that the proof by induction indeed works?? Proof (by contradiction) Assume that for some values of n, P(n) is false. Let be the least such n that P (n) is false. Cannot be 0, because P(0) is true. Thus, must be in the form = 1 +. Since < then by P ( is true. Therefore, by inductive hypothesis P ( + 1) must be true. It follows then that P ( is true.

Prove for ≥ 1 1 x 1! + 2 x 2! + 3 x 3! + ... + n x n! = (n +

1)! - 1

This could be also written by using ∑ notation

Example:

ProofBase case: n + 1

The left hand side is 1x1! The right hand side is 2! - 1. They are equal.

Inductive hypothesis: Suppose this holds 

We need to prove

Consider the left hand side

= (n+1)! – 1 + (n+1) x (n+1) = (n+1)! (1+n+1) -1 = (n+2)! -1

We can picture each proposition as a domino:

Suppose we have a sequence of propositions which we would like to prove:

P (0), P (1), P (2), P (3)… P (n)

P (k)P (k)

…..

A sequence of propositions is visualized as a sequence of dominos.

P (0)P (0) P (1)P (1) P (2)P (2) P (k)P (k) P (k+1)P (k+1)

When the domino falls (to right), the corresponding proposition is considered true:

P (k) true

P (k) true

Suppose that the dominos satisfy two constraints.Well positioned: if any domino falls to right, the next domino to right must fall also.

P (k) true

P (k) true

P (k+1) true

P (k+1) true

First domino has fallen to right.

P (0) true

P (0) true

…..

Than can conclude that all the domino fall.

P (0)P (0) P (1)P (1) P (2)P (2) P (k)P (k) P (k+1)P (k+1)

P (0)true

P (0)true

P (1)P (1) P (2)P (2) P (k)P (k) P (k+1)P (k+1)

…..

P (0) truetrueP (0) truetrue

P (1)true

P (1)true

P (2)P (2) P (k)P (k) P (k+1)P (k+1)

…..

P (0)

trueP (0)

true

P (1) true

P (1) true

P (2)True

P (2)True

P (k)P (k) P (k+1)P (k+1)

…..

P (0)

trueP (0)

true

P (1)

trueP (1)

true

P (2) trueP (2) true

truetrue

P (k)P (k) P (k+1)P (k+1)

…..

P (0) true

P (0) true

P (1) true

P (1) true

P (2) trueP (2) true

truetrue

P (k)true

P (k)true

P (k+1)P (k+1)

…..

P (0)

trueP (0)

true

P (1) true

P (1) true

P (2) true

P (2) true truetrue

P (k) true

P (k) true

P (k+1)

trueP (k+1)

true

Principle of Mathematical Induction.If:a) Basis P(1) is trueb) Induction ∀ n P(k) ® P(k+1) is true

P (0) true

P (0) true

P (1) true

P (1) true

P (2) true

P (2) true

truetrue

P (k) true

P (k) true

P (k+1)

trueP (k+1)

true

Then, ∀ n P (n) is true.  Use induction to prove that the sum of the first n odd integers is Base case (n=1): the sum of the first 1 odd integer is Assume p (k): the sum of the first k odd integers is Where, 1 + 3 + … + (2k+1) = Prove that

1 + 3 + … + + (2k-1) + (2k+1) = 1 + 3 + … + + (2k-1) + (2k+1) =

=

Yeah!

= 1

Prove that: 1 x 1! + 2 x 2! + … + n x n! = (n+1)! -1, ∀ nBase case (n=1): 1 x 1! = (1 x 1)! -1?Assume P (k 1 x 1! + 2 x 2! + … + k x k! = (k+1)! -1Prove that: x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+2)! -1 1 x 1! + 2 x 2! + … + k x k! + (k+1) (k+1)! = (k+1)! -1+ (k+1) (k+1)

= (1 + (k+1)) (k+1)! – 1 = (k+2) (k+1)! -1

= (k+2)! - 1

1 x 1! = 1

2! – 1 = 1