CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 4.0 International License. Based on a work at Permissions beyond the scope of this license may be available at LeeCreative Commons Attribution- NonCommercial-ShareAlike 4.0 International Licensehttp://peerinstruction4cs.org

Today’s Topics: 1. Quick wrap-up of Monday’s coin example 2. Strong vs regular induction 3. Strong induction examples:  Divisibility by a prime  Recursion sequence: product of fractions 2

Thm: For all prices p >= 8 cents, the price p can be paid using only 5-cent and 3-cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for p=8 (by example, e.g. p=3+5) Inductive step: Assume [or “Suppose”] that the theorem holds for some p  8. WTS that the theorem holds for p+1. p  8. So the inductive step holds, completing the proof. 3 Assume that p=5n+3m where n,m  0 are integers. We need to show that p+1=5a+3b for integers a,b  0. Partition to cases: Case I: n  1. In this case, p+1=5*(n-1)+3*(m+2). Case II: m  3. In this case, p+1=5*(n+2)+3*(m-3). Case III: n=0 and m  2. Then p=5n+3m  6 which is a contradiction to p  8.

We created an algorithm!  Our proof actually allows us to algorithmically find a way to pay p using 3-cent and 5-cent coins  Algorithm for price p:  start with 8=3+5  For x=8...p, in each step adjust the number of coins according to the modification rules we’ve constructed to maintain price x 4

Algorithm pseudo-code PayWithThreeCentsAndFiveCents: Input: price p  8. Output: integers n,m  0 so that p=5n+3m 1. Let x=8, n=1, m=1 (so that x=5n+3m). 2. While x<p: a) x:=x+1 b) If n  1, set n:=n-1, m:=m+2 c) Otherwise, set n:=n+2, m:=m-3 3. Return (n,m) 5

Algorithm pseudo-code PayWithThreeCentsAndFiveCents: Input: price p  8. Output: integers n,m  0 so that p=5n+3m 1. Let x=8, n=1, m=1 (so that x=5n+3m). 2. While x<p: a) x:=x+1 b) If n  1, set n:=n-1, m:=m+2 c) Otherwise, set n:=n+2, m:=m-3 3. Return (n,m) 6 Invariant: x=5n+3m We proved that n,m  0 in this process always; this is not immediate from the algorithm code

Algorithm run example  8=  9=  10 =  11=  12 = 7 x=8: n=1, m=1 While x<p: a)x:=x+1 b)If n  1, set n:=n-1, m:=m+2 c)Otherwise, set n:=n+2, m:=m-3 Invariant: x=5n+3m

Algorithm properties  Theorem: Algorithm uses at most two nickels (i.e n  2)  Proof: by induction on p  Try to prove it yourself first! 8 x=8: n=1, m=1 While x<p: a)x:=x+1 b)If n  1, set n:=n-1, m:=m+2 c)Otherwise, set n:=n+2, m:=m-3 Invariant: x=5n+3m

Algorithm properties  Theorem: Algorithm uses at most two nickels (i.e n  2). Proof: by induction on p Base case: p=8. Algorithm outputs n=m=1. Inductive hypothesis: p=5n+3m where n  2. WTS p+1=5a+3b where a  2. Proof by cases:  Case I: n  1. So p+1=5(n-1)+3(m+2) and a=n-1  2.  Case II: n=0. So p+1=5*2+3(m-3). a=2. In both cases p+1=5a+3b where a  2. QED 9 x=8: n=1, m=1 While x<p: a)x:=x+1 b)If n  1, set n:=n-1, m:=m+2 c)Otherwise, set n:=n+2, m:=m-3 Invariant: x=5n+3m

2. Strong induction examples DIVISIBILITY BY A PRIME 10

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) 11 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  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! 12

3. Strong induction examples DIVISIBILITY BY A PRIME 13

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). 14

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 15

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. 16

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

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. 18 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 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 For some integer n>1, k is divisible by a prime number, for all integers k where 2  k  n. WTS that So the inductive step holds, completing the proof. 19 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 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. 20

2. Strong induction examples RECURSION SEQUENCE: PRODUCT OF FRACTIONS 21

Definitions and properties for this proof  Product less than one:  Algebra, etc., as usual 22

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. 23

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. 24 A.0 B.1 C.2 D.3 E.Other/none/more than one

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. 25 A.For some int n>2, 0<d n <1. B.For some int n>2, 0<d k <1, for all integers k where 3  k  n. C.Other/none/more than one

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 theorem holds for n  2 WTS that So the inductive step holds, completing the proof. 26 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

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 for some int n  3, the theorem holds for all int k, n  k  3. (i.e. 0<d k <1 for all k between 3 and n, inclusive) 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. 27

3. Fibonacci numbers Verifying a solution 28

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! 29

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: 30 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  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 31

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… 32 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

Fibonacci numbers  Inductive step: need to show,  What can we use?  Definition of F n :  Inductive hypothesis:  That is, we need to show that 33

Fibonacci numbers  Finishing the inductive step.  Need to show:  Simplifying, need to show:  Choice of actually satisfied (this is why we chose it!) QED 34

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 35