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

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## Lecture 23 - Strong Mathematical Induction

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

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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 k k 1 k k 1 k 1 k 1 Example The Fibonacci Sequence Therefore the statement is true for all n 1 Example Trees A tree is a connected graph that contains no cycles Example Trees Notice that if you add an edge to a tree you necessarily create a cycle Also notice that if you delete an edge from a tree the graph becomes disconnected This suggests that a tree contains exactly the right number of edges Example Trees Theorem For all n 1 a tree with n nodes contains exactly n 1 edges Proof Base case Let n 1 Clearly there are 0 edges Example Trees Inductive case Suppose that the statement is true for all n k for some k 1 Let T be a tree with k 1 nodes Let v be an arbitrary node in T and let s be the index of v Let u1 u2 us be the adjacent nodes Delete the edges from v to u1 us Example Trees Then each ui is a node of a separate tree Ti Let ni be the number of nodes in tree Ti Note that n1 ns k By induction each tree Ti contains ni 1 edges Thus the total number of edges in tree T is n1 1 ns 1 s which equals k Example Trees Example Trees u2 v u1 u3 Example Trees u2 v u1 u3 Example Trees u2 v u1 4 nodes 3 edges u3 Example Trees 1 node 0 edges u2 v u1 4 nodes 3 edges u3 Example Trees 1 node 0 edges u2 v u3 u1 4 nodes 3 edges 3 nodes 2 edges Example Trees 1 node 0 edges u2 v u3 u1 4 nodes 3 edges 3 edges 3 nodes 2 edges Example Trees u2 v u3 u1 Total 3 0 2 3 8 edges Example Trees Example Trees u2 u1 v u3 Example Trees u2 u1 v u3 Example Trees u2 u1 6 nodes 5 edges v u3 Example Trees 1 node 0 edges u2 u1 6 nodes 5 edges v u3 Example Trees 1 node 0 edges u2 u1 6 nodes 5 edges v u3 1 nodes 0 edges Example Trees 1 node 0 edges u2 u1 6 nodes 5 edges v u3 3 edges 1 nodes 0 edges Example Trees u2 u1 v u3 Total 5 0 0 3 8 edges Example Binary Trees A binary tree structure consists of a set of nodes each with two pointers The left pointer points to the left child The right pointer points to the right child If there is no child then the pointer is null Example Binary Trees Corollary In a binary tree structure if there are n nodes then there are exactly n 1 null pointers Balanced Binary Strings Let S be the set of all finite binary strings that contain an equal number of 0 s and 1 s Define a set T as follows The empty string is in T If s T then 0s1 T and 1s0 T If s T and t T then st T Balanced Binary Strings Theorem T S Theorem T S The Paradox of the Pop Quiz A professor announces that one day during the semester he will give a pop quiz For the quiz to be a pop quiz it must be unexpected The Paradox of the Pop Quiz Theorem For all n 0 the professor cannot give the pop quiz n days before the last day of the semester The Paradox of the Pop Quiz Proof Base case n 0 He cannot give the pop quiz on the last day of the semester because everyone will be expecting it then The Paradox of the Pop Quiz Inductive step Suppose he cannot give it on any of the days 0 1 2 k days before the end of the semester for some k 0 Knowing this the students would be expecting it on the day k 1 days before the end of the semester So he can t give it on that day The Paradox of the Pop Quiz strong induction for all n 0 he cannot give the pop quiz n days before the end of the semester By

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