1. CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett

2. Today’s Topics: 1. Strong vs regular indiction 2. Strong induction examples:  Divisibility by a prime  Recursion sequence: product of fractions 2

3. 1. Strong induction examples DIVISIBILITY BY A PRIME 3

4. Strong vs regular induction  Prove:  n  1 P(n)  Base case: P(1)  Regular induction: P(n)  P(n+1)  Strong induction: (P(1)  …  P(n))  P(n+1)  Can use more assumptions to prove P(n+1) 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  Theorem: For all prices p >= 8 cents, the price p can be paid using only 5-cent and 3-cent coins  Proof:  Base case: 8=3+5, 9=3+3+3, 10=5+5  Assume it holds for all prices 1..p-1, prove for price p when p  11  Proof: since p-3  8 we can use the inductive hypothesis for p-3. To get price p simply add another 3-cent coin.  Much easier than standard induction! 5

6. 2. Strong induction examples DIVISIBILITY BY A PRIME 6

7. Definitions and properties for this proof  Definitions:  n is prime if  n is composite if n=ab for some 1<a,b<n  Prime or Composite exclusivity:  All integers greater than 1 are either prime or composite (exclusive or—can’t be both).  Definition of divisible:  n is divisible by d iff n = dk for some integer k.  2 is prime (you may assume this; it also follows from the definition). 7

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 8

9. Thm: For all integers n greater than 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. 9

10. Thm: For all integers n greater than 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 A.0 B.1 C.2 D.3 E.Other/none/more than one

11. Thm: For all integers n greater than 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. 11 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 greater than 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. 12 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 greater than 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<a,b<n+1. By the induction hypothesis, since a  n there exists a prime p which divides a. So p|a and a|n+1. We’ve already seen that this Implies that p|n+1 (in exam – give full details!) So the inductive step holds, completing the proof. 13

14. 2. Strong induction examples RECURSION SEQUENCE: PRODUCT OF FRACTIONS 14

15. Definitions and properties for this proof  Product less than one:  Algebra, etc., as usual 15

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<d n <1. 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. 16

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<d n <1. 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. 17 A.0 B.1 C.2 D.3 E.Other/none/more than one

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<d n <1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _ 1,2 _____. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. 18 A.For some int n>0, 0<d n <1. B.For some int n>1, 0<d k <1, for all integers k where 1  k  n. C.Other/none/more than one

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<d n <1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _ 1,2 _____. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. 19 A.For some int n>0, 0<d n <1. B.For some int n>1, 0<d k <1, for all integers k where 1  k  n. C.0<d n+1 <1 D.Other/none/more than one

20. 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<d n <1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n=1,2. Inductive step: Assume [or “Suppose”] that the theorem holds for n  2. WTS that 0<d n+1 <1. By definition, d n+1 =d n d n-1. By the inductive hypothesis, 0<d n-1 <1 and 0<d n <1. Hence, 0<d n+1 <1. So the inductive step holds, completing the proof. 20

21. 3. Fibonacci numbers Verifying a solution 21

22. 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?  YES! 22

23. Fibonacci numbers  Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1.  We will prove an upper bound:  Proof by strong induction.  Base case: 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  Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1.  We will prove an upper bound:  Proof by strong induction.  Base case: 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  Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1.  We will prove an upper bound:  Proof by strong induction.  Base case: n=1, n=2. Verify by direct calculation 25

26. Fibonacci numbers  Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1.  Theorem:  Base cases: n=1,n=2  Inductive step: show… 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. Fibonacci numbers  Rule: F 1 =1, F 2 =1, F n =F n-2 +F n-1.  Theorem:  Base cases: n=1,n=2  Inductive step: show… 27 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

28. Fibonacci numbers  Inductive step: need to show,  What can we use?  Definition of F n :  Inductive hypothesis:  That is, we need to show that 28

29. Fibonacci numbers  Finishing the inductive step.  Need to show:  Simplifying, need to show:  Choice of actually satisfied (this is why we chose it!) QED 29

30. 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 30