# CSE 20 DISCRETE MATH Prof. Shachar Lovett Clicker frequency: CA.

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

Todays topics Strong induction Section 3.6.1 in Jenkyns, Stephenson

Strong vs regular induction

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

Example for the power of strong induction 5

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

Definitions and properties for this proof

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

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.

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

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

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

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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_13.jpg", "name": "Thm: For all integers n>1, n is divisible by a prime number.", "description": "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

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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_15.jpg", "name": "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

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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_16.jpg", "name": "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

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, 00, 01, 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_17.jpg", "name": "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, 00, 01, 0

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, 00, 01, 0 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_18.jpg", "name": "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, 00, 01, 0

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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4317641/slides/slide_19.jpg", "name": "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

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?

Fibonacci numbers

22

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

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

Fibonacci numbers 25

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

Proof of inductive case

Proof of inductive case (contd)

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

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

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