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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