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Induction

# Induction - CSE 1400 Applied Discrete Mathematics...

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CSE 1400 Applied Discrete Mathematics Mathematical Induction Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Induction 1 Summations 1 The Tower of Hanoi and Mersenne Numbers 7 Rabbit Population Growth and Fibonacci Numbers 8 Induction on Arithmetic and Geometric Sequences 10 Problems on Induction 13 Abstract The word induction has many meanings. For us it is a formal proof process that a predicate p ( n ) is True for all natural numbers n be- longing to some set, most often the set of natural numbers N = { 0, 1, 2, . . . } . The principle of induction is: If set X contains zero and if x X implies its successor x + 1 X , then every natural number is in X , that is X = N . Induction Induction often involves sums: If the sum of n term equals some formula, then the sum of n + 1 terms equals the same formula with appropriate substitutions. Induction often involves recurrsion: A function f ( n ) satisfies a recurrence relation. An inductive proof of the truth of predicate p ( n ) for all natural numbers n has two steps. 1 . A proof that p ( 0 ) is True . 2 . A proof of the truth of the condi- tional statement If p ( n ) is True , then p ( n + 1 ) is True . A short notation for this is p ( 0 ) p ( n ) p ( n + 1 ) ( n N ) p ( n ) Which is read " p ( 0 ) is True ; and if p ( n ) is True , then p ( n + 1 ) is True . Therefore, p ( n ) is True for all natural numbers n ." Summations Consider the predicate: The sum of the first n natural numbers is n ( n - 1 ) /2.

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cse 1400 applied discrete mathematics mathematical induction 2 You could use the notation p ( n ) to denote the truth value of the statement “The sum of the first n natu- ral numbers is n ( n - 1 ) /2.” For instance, the sum of the first 5 natural numbers is 0 + 1 + 2 + 3 + 4 = 10 = 5 · ( 5 - 1 ) /2 = 5 · 4/2 Since the sum of the first 5 natural numbers is 10, the sum of the first 6 natural numbers is 10 + 5 = 15 = 6 × 5 2 The heart of mathematical induction relies on ( 1 ) noticing a pattern for small initial cases, and ( 2 ) establishing a link allowing a proof of of an instance from the truth of previous instances. In this case, the pattern is The sum of n terms in an arithmetic progression is n times the average of the first and last term. 0 + 1 + 2 + 3 + 4 + · · · + ( n - 1 ) = n × n - 1 2 which can be phrases as “the sum of the first n natural numbers is n times n - 1 divided by 2.” or p ( n ) for short. To establish the link between the truth of p ( n ) and p ( n + 1 ) re- quires a little algebra. We must generalize the instance above where the step was from 5 to 6. If the sum of the first n natural numbers The sum of the first n + 1 natural numbers is the sum of the first n natural numbers plus n , the ( n + 1 ) st natural number. is n ( n - 1 ) /2, then the sum of the first n + 1 natural numbers is n ( n - 1 ) /2 plus n . That is, if 0 + 1 + 2 + 3 + 4 + · · · + ( n - 1 ) = n ( n - 1 ) 2 then [ 0 + 1 + 2 + · · · + ( n - 1 )] + n = n ( n - 1 ) 2 + n = n n - 1 2 + 1 = ( n + 1 ) n 2 This shows that if the function n ( n - 1 ) /2 correctly computes the value in the n th case, then it correctly computes the value in the ( n + 1 ) st case.
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Induction - CSE 1400 Applied Discrete Mathematics...

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