Download presentation

Presentation is loading. Please wait.

Published byMalcolm Young Modified about 1 year ago

1
CSE 20 DISCRETE MATH Prof. Shachar Lovett http://cseweb.ucsd.edu/classes/wi15/cse20-a/ Clicker frequency: CA

2
Todays topics Strong induction Section 3.6.1 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/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

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

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

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, 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, 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

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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on institute management system Ppt on solid dielectrics industries Ppt on earth and space issues Ppt on culture and science in the ancient period Free download ppt on human evolution Ppt on supply chain management of a mule Surface conduction electron emitter display ppt online Ppt on varactor diode circuit Ppt on world stock exchange Download ppt on chemical bonding and molecular structure