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CSE 20 DISCRETE MATH Prof. Shachar Lovett http://cseweb.ucsd.edu/classes/wi15/cse20-a/ Clicker frequency: CA

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Todays topics Strong induction Section 3.6.1 in Jenkyns, Stephenson

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Strong vs regular induction

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P(1)P(2)P(3)P(n)P(n+1) …… P(1)P(2)P(3)P(n)P(n+1) ……

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Example for the power of strong induction 5

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Another example: divisibility Thm: For all integers n>1, n is divisible by a prime number. Before proving it (using strong induction), lets first review some of the basic definitions, but now make them precise

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Definitions and properties for this proof

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Definitions and properties for this proof (cont.) Goes without saying at this point: The set of Integers is closed under addition and multiplication Use algebra as needed

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Thm: For all integers n>1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof.

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Thm: For all integers n>1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. A.0 B.1 C.2 D.3 E.Other/none/more than one

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Thm: For all integers n>1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _2______. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. A.For some integer n>1, n is divisible by a prime number. B.For some integer n>1, k is divisible by a prime number, for all integers k where 2 k n. C.Other/none/more than one

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Thm: For all integers n>1, n is divisible by a prime number. A.n+1 is divisible by a prime number. B.k+1 is divisible by a prime number. C.Other/none/more than one

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Thm: For all integers n>1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n=2. Inductive step: Assume that for some n 2, all integers 2 k n are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1
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Theorems about recursive definitions

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Definition of the sequence: d 1 = 9/10 d 2 = 10/11 d k = (d k-1 )(d k-2 ) for all integers k 3 Thm: For all integers n>0, 0

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Definition of the sequence: d 1 = 9/10 d 2 = 10/11 d k = (d k-1 )(d k-2 ) for all integers k 3 Thm: For all integers n>0, 0

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Definition of the sequence: d 1 = 9/10 d 2 = 10/11 d k = (d k-1 )(d k-2 ) for all integers k 3 Thm: For all integers n>0, 0

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Definition of the sequence: d 1 = 9/10 d 2 = 10/11 d k = (d k-1 )(d k-2 ) for all integers k 3 Thm: For all integers n>0, 0

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Definition of the sequence: d 1 = 9/10 d 2 = 10/11 d k = (d k-1 )(d k-2 ) for all integers k 3 Thm: For all integers n>0, 0

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Fibonacci numbers 1,1,2,3,5,8,13,21,… Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1. Question: can we derive an expression for the n-th term?

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Fibonacci numbers

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Fibonacci numbers 23 A.n=1 B.n=2 C.n=1 and n=2 D.n=1 and n=2 and n=3 E.Other

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Fibonacci numbers 24 A.n=1 B.n=2 C.n=1 and n=2 D.n=1 and n=2 and n=3 E.Other

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Fibonacci numbers 25

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Fibonacci numbers 26 A.F n =F n-1 +F n-2 B.F n F n-1 +F n-2 C.F n =r n D.F n r n E.Other

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Proof of inductive case

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Proof of inductive case (contd)

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Fibonacci numbers - recap Recursive definition of a sequence Base case: verify for n=1, n=2 Inductive step: Formulated what needed to be shown as an algebraic inequality, using the definition of F n and the inductive hypothesis Simplified algebraic inequality Proved the simplified version 29

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Next class Applying proof techniques to analyze algorithms Read section 3.7 in Jenkyns, Stephenson

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