09 Mathematical Induction

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    Mathematical Induction

    CS 202

    Epp, chapter 4Aaron Bloomfield

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    How do you climb infinite stairs?

    Not a rhetorical question!

    First, you get to the base platform of thestaircase

    Then repeat:

    From your current position, move one step up

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    Lets use that as a proof method

    First, show P(x) is true for x=0 This is the base of the stairs

    Then, show that if its true for some value n, then it istrue for n+1 Show: P(n) P(n+1)

    This is climbing the stairs Let n=0. Since its true for P(0) (base case), its true for n=1 Let n=1. Since its true for P(1) (previous bullet), its true for n=2 Let n=2. Since its true for P(2) (previous bullet), its true for n=3 Let n=3

    And onwards to infinity

    Thus, we have shown it to be true for allnon-negativenumbers

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    What is induction?

    A method of proof It does not generate answers:

    it only can prove them

    Three parts: Base case(s): show it is true

    for one element (get to the stairs base platform)

    Inductive hypothesis: assumeit is true for any given element

    (assume you are on a stair) Must be clearly labeled!!!

    Show that if it true for the nexthighest element

    (show you can move to the nextstair)

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    Show that the sum of the first n oddintegers is n2

    Example: If n= 5, 1+3+5+7+9 = 25 = 52

    Formally, show:

    Base case: Show that P(1) is true

    Induction example

    n

    i

    ni

    1

    212

    11

    11)(21

    1

    2

    i

    i

    )P(where)P( nnn

    )1P(

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    Inductive hypothesis: assume true for k

    Thus, we assume that P(k) is true, or that

    Note: we dont yet know if this is true or not!

    Inductive step: show true for k+1 We want to show that:

    k

    i

    ki1

    212

    1

    1

    2)1(12

    k

    i

    ki

    Induction example, continued

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    12121)1(2

    2

    1

    kkik

    k

    i

    121)1(222

    kkkk

    Induction example, continued

    Recall the inductive hypothesis:

    Proof of inductive step:

    k

    i

    ki

    1

    212

    121222

    kkkk

    21

    1

    )1(12

    ki

    k

    i

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    What did we show

    Base case: P(1) If P(k) was true, then P(k+1) is true

    i.e., P(k) P(k+1)

    We know its true for P(1)

    Because of P(k) P(k+1), if its true for P(1), then its true for P(2) Because of P(k) P(k+1), if its true for P(2), then its true for P(3) Because of P(k) P(k+1), if its true for P(3), then its true for P(4)

    Because of P(k) P(k+1), if its true for P(4), then its true for P(5) And onwards to infinity

    Thus, it is true for all possible values of n

    In other words, we showed that:

    )P()1P()P()1P( nnkkk

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    The idea behind inductive proofs

    Show the base case

    Show the inductive hypothesis

    Manipulate the inductive step so that youcan substitute in part of the inductivehypothesis

    Show the inductive step

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    Second induction example

    Show the sum of the first npositive evenintegers is n2 + n

    Rephrased:

    The three parts:

    Base case

    Inductive hypothesis

    Inductive step

    n

    i

    nninnn

    1

    22)P(where)P(

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    Second induction example,continued

    Base case: Show P(1):

    Inductive hypothesis: Assume

    Inductive step: Show

    22

    11)(2)1P(1

    1

    2

    i

    i

    k

    i

    kkik

    1

    22)P(

    )1()1(2)1P(2

    1

    1

    kkik

    k

    i

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    232322

    kkkk

    1)1()1(222

    kkkkk

    1)1(2)1(22

    1

    kkik

    k

    i

    1)1(22

    1

    1

    kki

    k

    i

    Second induction example,continued

    Recall our inductivehypothesis:

    k

    i

    kkik

    1

    22)P(

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    Notes on proofs by induction

    We manipulate the k+1 case to make partof it look like the kcase

    We then replace that part with the otherside of the kcase

    k

    i

    kkik

    1

    22)P(

    232322

    kkkk

    1)1()1(222

    kkkkk

    1)1(2)1(22

    1

    kkik

    k

    i

    1)1(22

    1

    1

    kki

    k

    i

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    Third induction example

    Show

    Base case: n= 1

    Inductive hypothesis: assume

    6

    )12)(1(

    1

    2

    nnni

    n

    i

    11

    6

    61

    6)12)(11(1

    2

    1

    1

    2

    i

    i

    6

    )12)(1(

    1

    2

    kkki

    k

    i

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    Third induction example

    Inductive step: show6

    )1)1(2)(1)1)((1(1

    1

    2

    kkki

    k

    i

    6

    )32)(2)(1()1(

    1

    22

    kkkik

    k

    i

    61392613922323

    kkkkkk

    )32)(2)(1()12)(1()1(62

    kkkkkkk

    6)32)(2)(1(

    6)12)(1()1(

    2 kkkkkkk

    6

    )1)1(2)(1)1)((1(1

    1

    2

    kkki

    k

    i

    6

    )12)(1(

    1

    2

    kkk

    i

    k

    i

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    Third induction again: what if yourinductive hypothesis was wrong?

    Show:

    Base case: n= 1:

    But lets continue anyway

    Inductive hypothesis: assume

    6

    )22)(1(

    1

    2

    nnni

    n

    i

    6

    71

    6

    71

    6

    )22)(11(1

    2

    1

    1

    2

    i

    i

    6

    )22)(1(

    1

    2

    kkki

    k

    i

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    Third induction again: what if yourinductive hypothesis was wrong?

    Inductive step: show6

    )2)1(2)(1)1)((1(1

    1

    2

    kkki

    k

    i

    6

    )42)(2)(1()1(

    1

    22

    kkkik

    k

    i

    8161026141022323

    kkkkkk

    )42)(2)(1()22)(1()1(62

    kkkkkkk

    6)42)(2)(1(

    6)22)(1()1(

    2 kkkkkkk

    6

    )2)1(2)(1)1)((1(1

    1

    2

    kkki

    k

    i

    6

    )22)(1(

    1

    2

    kkk

    i

    k

    i

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    Fourth induction example

    S that n! < nn for all n> 1

    Base case: n= 22! < 22

    2 < 4

    Inductive hypothesis: assume k! < kk

    Inductive step: show that (k+1)! < (k+1)k+1

    )!1( k

    !)1( kk

    kkk)1(

    k

    kk)1)(1(

    1

    )1(

    kk

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    Strong induction

    Weak mathematical induction assumesP(k) is true, and uses that (and only that!)to show P(k+1) is true

    Strong mathematical induction assumesP(1), P(2), , P(k) are all true, and uses

    that to show that P(k+1) is true.

    )1P()P(...)3P()2P()1P( kk

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    Strong induction example 1

    Show that any number > 1 can be writtenas the product of one or more primes

    Base case: P(2) 2 is the product of 2 (remember that 1 is not

    prime!)

    Inductive hypothesis: assume P(2), P(3),, P(k) are all true

    Inductive step: Show that P(k+1) is true

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    Strong induction example 1

    Inductive step: Show that P(k+1) is true

    There are two cases:

    k+1 is prime

    It can then be written as the product of k+1

    k+1 is composite

    It can be written as the product of two composites,

    a and b, where 2 ab< k+1 By the inductive hypothesis, both P(a) and P(b) are

    true

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    Strong induction vs. non-stronginduction

    Determine which amounts of postage canbe written with 5 and 6 cent stamps

    Prove using both versions of induction

    Answer: any postage 20

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    Answer via mathematical induction

    Show base case: P(20):

    20 = 5 + 5 + 5 + 5

    Inductive hypothesis: Assume P(k) is true

    Inductive step: Show that P(k+1) is true If P(k) uses a 5 cent stamp, replace that stamp with a

    6 cent stamp

    If P(k) does not use a 5 cent stamp, it must use only 6

    cent stamps

    Since k> 18, there must be four 6 cent stamps

    Replace these with five 5 cent stamps to obtain k+1

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    Answer via strong induction

    Show base cases: P(20), P(21), P(22), P(23), and P(24)

    20 = 5 + 5 + 5 + 5

    21 = 5 + 5 + 5 + 6

    22 = 5 + 5 + 6 + 6

    23 = 5 + 6 + 6 + 6

    24 = 6 + 6 + 6 + 6

    Inductive hypothesis: Assume P

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