Slide 1Slide 2Slide 3Slide 4Standard NotationSlide 6The Natural NumbersPlato: The Domino Principle works for an infinite row of dominoesPlato’s Dominoes One for each natural numberSlide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 54COMPSCI 102Introduction to Discrete MathematicsLecture 4 (September 2, 2009)Inductive ReasoningDominoesDomino Principle: Line up any number of dominos in a row; knock the first one over and they will all fallDominoes Numbered 1 to nFk = “The kth domino falls”If we set them up in a row then each one is set up to knock over the next:For all 1 ≤ k < n:Fk Fk+1F1 F2 F3 …F1 All Dominoes FallStandard Notation“for all” is written “”Example:For all k>0, P(k) k>0, P(k)=Dominoes Numbered 0 to n-1Fk = “The kth domino falls”k, 0 ≤ k < n-1:Fk Fk+1F0 F1 F2 …F0 All Dominoes FallThe Natural NumbersOne domino for each natural number:0 1 2 3 …= { 0, 1, 2, 3, . . . }Plato: The Domino Principle works for an infinite row of dominoesAristotle: Never seen an infinite number of anything, much less dominoes.Plato’s DominoesOne for each natural numberTheorem: An infinite row of dominoes, one domino for each natural number.Knock over the first domino and they all will fallSuppose they don’t all fall. Let k > 0 be the lowest numbered domino that remains standing1. Domino k-1 ≥ 0 did fall, but k-1 will knock over domino k. Thus, domino k must fall and remain standing. Contradiction.Proof: 1Via law of excluded middle over the infinite and well-ordering of natural numbers.Mathematical Induction statements proved instead of dominoes fallenInfinite sequence ofdominoesInfinite sequence of statements: S0, S1, …Fk = “domino k fell” Fk = “Sk proved”Conclude that Fk is true for all kEstablish: 1. F02. For all k, Fk Fk+1Inductive ProofsTo Prove k , SkEstablish “Base Case”: S0Establish that k, Sk Sk+1k, Sk Sk+1Assume hypothetically that Sk for any particular k; Conclude that Sk+1Theorem?The sum of the first n odd numbers is n2Check on small values:1 = 11+3 = 41+3+5 = 91+3+5+7 = 16Theorem?The sum of the first n odd numbers is n2The kth odd number is (2k – 1), when k > 0Sn is the statement that: “1+3+5+(2k-1)+...+(2n-1) = n2”Sn = “1 + 3 + 5 + (2k-1) + . . +(2n-1) = n2”Establishing that n ≥ 1 SnBase Case: S1Assume “Induction Hypothesis”: SkThat means: 1+3+5+…+ (2k-1) = k21+3+5+…+ (2k-1)+(2k+1) = k2 +(2k+1)Sum of first k+1 odd numbers = (k+1)2Domino Property:TheoremThe sum of the first n odd numbers is n2Primes: Note: 1 is not considered primeA natural number n > 1 is a prime if it has no divisors besides 1 and itselfTheorem?Every natural number > 1 can be factored into primesNote: factorization is unique.“Fundamental Theorem of Arithmetic”“Unique-Prime-Factorization Theorem”Theorem?Every natural number > 1 can be factored into primesSn = “n can be factored into primes”Base case:2 is prime S2 is trueSk-1 = “k-1 can be factored into primes”How do we use the fact:Sk = “k can be factored into primes”to prove that:This shows a technical point about mathe-matical inductionTheorem?Every natural number > 1 can be factored into primesA different approach:Assume 2,3,…,k-1 all can be factored into primesThen show that k can be factored into primesAll Previous InductionTo Prove k, SkEstablish Base Case: S0Establish Domino Effect:Assume j<k, Sj use that to derive SkAll Previous InductionTo Prove k, SkEstablish Base Case: S0Establish Domino Effect:Assume j<k, Sj use that to derive SkAlso called “Strong Induction”Let k be any number“All Previous” InductionRepackaged AsStandard InductionEstablish Base Case: S0Establish Domino Effect:Let k be any numberAssume j<k, Sj Prove SkDefine Ti = j ≤ i, SjEstablish Base Case T0Establish that k, Tk Tk+1Assume Tk-1 Prove TkAnd there are more ways to do inductive proofsMethod of Infinite DescentShow that for any counter-example you find a smaller oneRene DescartesIf a counter-example exists there would be an infinite sequence of smaller and smaller counter examplesTheorem:Every natural number > 1 can be factored into primesLet n be a counter-exampleHence n is not prime, so n = abIf both a and b had prime factorizations, then n would tooThus a or b is a smaller counter-exampleInvariant (n): 1. Not varying; constant. 2. Mathematics. Unaffected by a designated operation, as a transformation of coordinates.Yet another way of packaging inductive reasoning is to define “invariants”Invariant (n): 3. Programming. A rule, such as the ordering of an ordered list, that applies throughout the life of a data structure or procedure. Each change to the data structure maintains the correctness of the invariantInvariant InductionSuppose we have a time varying world state: W0, W1, W2, …Argue that S is true of the initial worldShow that if S is true of some world – then S remains true after one permissible operation is performedEach state change is assumed to come from a list of permissible operations. We seek to prove that statement S is true of all future worldsOdd/Even Handshaking Theorem At any party at any point in time define a person’s parity as ODD/EVEN according to the number of hands they have shakenStatement: The number of people of odd parity must be evenIf 2 people of the same parity shake, they both change and hence the odd parity count changes by 2 – and remains evenStatement: The number of people of odd parity must be evenInitial case: Zero hands have been shaken at the start of a party, so zero people have odd parityInvariant Argument: If 2 people of different parities shake, then they both swap parities and the odd parity count is unchangedInductive reasoning is the high level idea“Standard” Induction“All Previous” Induction “Least Counter-example”“Invariants”all just different packagingInduction is also how we can define and construct our worldSo many things, from buildings to computers, are built up stage by stage, module by module, each depending on the previous stagesn 0 1 2 3 4 5 6 7F(n)Inductive
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