# H-SC MATH 262 - Lecture 23 - Strong Mathematical Induction (44 pages)

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## Lecture 23 - Strong Mathematical Induction

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## Lecture 23 - Strong Mathematical Induction

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Pages:
44
School:
Hampden-Sydney College
Course:
Math 262 - Discrete Mathematics
##### Discrete Mathematics Documents

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Strong Mathematical Induction Lecture 23 Section 4 4 Tue Feb 27 2007 Announcements Thursday at 4 00 cookies at 3 30 The Principle of Strong Mathematical Induction Let P n be a predicate defined for integers n Let a be an integer If the following two statements are true P a P a 1 P b For all integers k b if P a P a 1 P k then P k 1 then the statement For all integers n a P n is true The Principle of Strong Mathematical Induction The range a a 1 a 2 b represents the number of previous cases that the inductive step depends on Usually this is 1 We will see one example where it is 2 Example Factoring Integers Theorem Every integer n 2 factors into primes Proof Base case Let n 2 Then n is already a prime so the statement is true Example Factoring Integers Inductive case Suppose the statement is true for all integers from 2 to k for some k 2 Consider k 1 Either k 1 is prime or it is not prime If it is prime then we are done If it is not prime then it factors as a b for some integers a b with a b 2 Example Factoring Integers By the induction hypothesis a and b themselves factor into the product of primes Therefore k 1 factors into the product of primes Example The Fibonacci Sequence The Fibonacci sequence fn is defined by f0 0 f1 1 fn fn 1 fn 2 for all n 2 The first few terms are 0 1 1 2 3 5 8 13 21 34 Example The Fibonacci Sequence Theorem Let 1 5 1 5 and 2 2 Then for all n 0 n n fn Example The Fibonacci Sequence Proof Base cases n 0 Then 0 0 0 Let n 1 Then 1 1 1 Let Inductive case Suppose the equation holds for all n from 0 to k for some k 1 Consider fk 1 A Lemma Lemma n 1 n n 1 and n 1 n n 1 for all n 1 Proof for Base case Let n 1 Then a direct calculation shows that 2 1 1 0 A Lemma Inductive step Now suppose it is true for some integer k 1 That is k 1 k k 1 Multiply the equation through by The result is k 2 k 1 k So the statement is true when n k 1 Therefore the statement is true for all n 1 Example The Fibonacci Sequence Back to our theorem f k 1 f k f k 1 k k k 1 k 1

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