CSE 20 – Discrete Mathematics - PowerPoint Presentation

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CSE 20 – Discrete Mathematics

Dr. Cynthia Bailey LeeDr. Shachar Lovett

 

                         

Peer Instruction in Discrete Mathematics by 

Cynthia

Lee

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Today’s Topics:

Quick wrap-up of Monday’s coin exampleStrong vs regular inductionStrong induction examples:Divisibility by a primeRecursion sequence: product of fractions

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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.

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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.

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We created an algorithm!

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

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Algorithm pseudo-code

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

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Algorithm pseudo-code

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

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Invariant: x=5n+3m

Invariant: x=5n+3m

We proved that n,m

0 in this process always; this is not immediate from the algorithm code

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Algorithm run example

8=9=10 =11=12 =

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x=8: n=1, m=1While x<p:x:=x+1If n1, set n:=n-1, m:=m+2Otherwise, set n:=n+2, m:=m-3

Invariant: x=5n+3m

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Algorithm properties

Theorem: Algorithm uses at most two nickels (i.e n2)Proof: by induction on pTry to prove it yourself first!

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x=8: n=1, m=1While x<p:x:=x+1If n1, set n:=n-1, m:=m+2Otherwise, set n:=n+2, m:=m-3

Invariant: x=5n+3m

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Algorithm properties

Theorem: Algorithm uses at most two nickels (i.e n2).Proof: by induction on pBase 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

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x=8: n=1, m=1While x<p:x:=x+1If n1, set n:=n-1, m:=m+2Otherwise, set n:=n+2, m:=m-3

Invariant: x=5n+3m

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2. Strong induction examples

DIVISIBILITY BY A PRIME

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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)

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P(1)

P(2)

P(3)

P(n)

P(n+1)

…

…

P(1)

P(2)

P(3)

P(n)

P(n+1)

…

…

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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 coinsProof:Base case: 8=3+5, 9=3+3+3, 10=5+5Assume it holds for all prices 1..p-1, prove for price p when p11Proof: 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!

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3. Strong induction examples

DIVISIBILITY BY A PRIME

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Definitions and properties for this proof

Definitions:n is prime if n is composite if n=ab for some 1<a,b<nPrime 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).

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Definitions and properties for this proof (cont.)

Goes without saying at this point: The set of Integers is closed under addition and multiplicationUse algebra as needed

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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”] thatWTS that So the inductive step holds, completing the proof.

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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”] thatWTS that So the inductive step holds, completing the proof.

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0

1

2

3

Other/none/more than one

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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”] thatWTS that So the inductive step holds, completing the proof.

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For some integer n>1, n is divisible by a prime number.

For some integer n>1, k is divisible by a prime number, for all integers k where 2



k



n.

Other/none/more than one

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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”] thatFor 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.

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n+1 is divisible by a prime number.

k+1 is divisible by a prime number.

Other/none/more than one

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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.

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2. Strong induction examples

RECURSION SEQUENCE: PRODUCT OF FRACTIONS

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Definitions and properties for this proof

Product less than one:Algebra, etc., as usual

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Definition of the sequence:d1 = 9/10d2 = 10/11dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1.

Proof (by strong mathematical induction):Basis step: Show the theorem holds for n = ________. Inductive step:Assume [or “Suppose”] thatWTS thatSo the inductive step holds, completing the proof.

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Definition of the sequence:d1 = 9/10d2 = 10/11dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1.

Proof (by strong mathematical induction):Basis step: Show the theorem holds for n = ________. Inductive step:Assume [or “Suppose”] thatWTS thatSo the inductive step holds, completing the proof.

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0

1

2

3

Other/none/more than one

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Definition of the sequence:d1 = 9/10d2 = 10/11dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1.

Proof (by strong mathematical induction):Basis step: Show the theorem holds for n = 1,2. Inductive step:Assume [or “Suppose”] thatWTS thatSo the inductive step holds, completing the proof.

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For some

int

n>2, 0<

d

n

<

1

.

For some

int

n>2, 0<

d

k

<

1,

for all integers k where

3



k



n

.

Other/none/more than one

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Definition of the sequence:d1 = 9/10d2 = 10/11dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1.

Proof (by strong mathematical induction):Basis step: Show the theorem holds for n = ________. Inductive step:Assume [or “Suppose”] thattheorem holds for n2WTS thatSo the inductive step holds, completing the proof.

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For some

int

n>0

,

0<

d

n

<

1

.

For some

int

n>1,

0<

d

k

<

1,

for all integers k where

1



k



n.

0<d

n+1

<1

Other/none/more than one

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Definition of the sequence:d1 = 9/10d2 = 10/11dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<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<dk<1 for all k between 3 and n, inclusive)WTS that 0<dn+1<1.By definition, dn+1=dn dn-1. By the inductive hypothesis, 0<dn-1<1 and 0<dn<1. Hence, 0<dn+1<1.So the inductive step holds, completing the proof.

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3. Fibonacci numbers

Verifying a solution

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Fibonacci numbers

1,1,2,3,5,8,13,21,…Rule: F1=1, F2=1, Fn=Fn-2+Fn-1.Question: can we derive an expression for the n-th term?YES!

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Fibonacci numbers

Rule: F1=1, F2=1, Fn=Fn-2+Fn-1.We will prove an upper bound:Proof by strong induction.Base case:

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n=1n=2n=1 and n=2n=1 and n=2 and n=3Other

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Fibonacci numbers

Rule: F1=1, F2=1, Fn=Fn-2+Fn-1.We will prove an upper bound:Proof by strong induction.Base case: n=1, n=2. Verify by direct calculation

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Fibonacci numbers

Rule: F1=1, F2=1, Fn=Fn-2+Fn-1.Theorem:Base cases: n=1,n=2Inductive step: show…

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Fn=Fn-1+Fn-2FnFn-1+Fn-2Fn=rnFn rnOther

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Fibonacci numbers

Inductive step: need to show , What can we use?Definition of Fn:Inductive hypothesis:That is, we need to show that

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Fibonacci numbers

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

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Fibonacci numbers - recap

Recursive definition of a sequenceBase case: verify for n=1, n-2Inductive step:Formulated what needed to be shown as an algebraic inequality, using the definition of Fn and the inductive hypothesisSimplified algebraic inequalityProved the simplified version

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