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

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Presentation on theme: "CSE 20 DISCRETE MATH Prof. Shachar Lovett Clicker frequency: CA."— Presentation transcript:

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

2 Todays topics Strong induction Section in Jenkyns, Stephenson

3 Strong vs regular induction

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

5 Example for the power of strong induction 5

6 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

7 Definitions and properties for this proof

8 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

9 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.

10 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

11 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

12 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

13 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/4317641/14/slides/slide_12.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

14 Theorems about recursive definitions

15 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/4317641/14/slides/slide_14.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

16 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/4317641/14/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

17 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/4317641/14/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, 00, 01, 0

18 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/4317641/14/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

19 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/4317641/14/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, 0

20 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?

21 Fibonacci numbers

22 22

23 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

24 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

25 Fibonacci numbers 25

26 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

27 Proof of inductive case

28 Proof of inductive case (contd)

29 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

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


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