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Mathematical Induction I

Lecture 4: Sep 16

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Proving For-All Statements

Objective: +rove

It is ver common to prove statements of this form. Some ,*amples:

-or an odd number m' mi is odd for all nonne#ative inte#er i.

)n inte#er n / 1 is divisible b a prime number.

&0auchSchwar inequalit( -or an a1'2'an' and an b1'2bn

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Universal Generalization

( )

. ( )

A R c

A x R x

→

→∀valid rule providin# c is independent of )

3ne wa to prove a forall statement is to prove that &c( is true for an c'but this is often difficult to prove directl

&e.#. consider the statements in the previous slide(.

Mathematical induction provides another wa to prove a forall statement.

It allows us to prove the statement step-by-step.

Let us first see the idea in two e*amples.

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O Po!ers Are O

-act: If m is odd and n is odd' then nm is odd.

+roposition: for an odd number m' mi is odd for all nonne#ative inte#er i.

Let +&i( be the proposition that m i is odd.

$ +&1( is true b definition.$ +&5( is true b +&1( and the fact.

$ +&( is true b +&5( and the fact.

$ +&i71( is true b +&i( and the fact.

$ So +&i( is true for all i.

Idea of induction.

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"ivisibility by a Prime

Theorem# )n inte#er n / 1 is divisible b a prime number.

Idea of induction.

$Let n be an inte#er.

$If n is a prime number' then we are done.

$3therwise' n 8 ab' both are smaller than n.

$If a or b is a prime number' then we are done.

$3therwise' a 8 cd' both are smaller than a.

$If c or d is a prime number' then we are done.

$3therwise' repeat this ar#ument' since the numbers are

#ettin# smaller and smaller' this will eventuall stop and

we have found a prime factor of n.

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3b9ective: +rove

$ea o% $nuction

This is to prove

The idea of induction is to first prove +&( unconditionall'

then use +&( to prove +&1(

then use +&1( to prove +&5(

and repeat this to infinit2

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The $nuction &ule

and &from n to n 71('

proves ' 1' 5' '2.

+ &(' ∀n ∈; + &n (+ &n 71(

∀m ∈; + &m (

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This Lecture

$ The idea of mathematical induction

$ %asic induction proofs &e.#. equalit' inequalit' propert'etc(

$ Inductive constructions

$ ) parado*

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Proving an '(uality

Let +&n( be the induction hpothesis that the statement is true for n.

%ase case: +&1( is true

Induction step: assume +&n( is true' prove +&n71( is true.

because both L>S and >S equal to 1

That is' assumin#:

?ant to prove:

This is much easier to prove than provin# it directl'

because we alread =now the sum of the first n terms@

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Proving an '(uality




b induction

because both L>S and >S equal to 1

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Proving an '(uality




b induction

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Proving a Property

%ase 0ase &n 8 1(:

Induction Step: )ssume + &i ( for some i ≥ 1 and prove + &i 7 1(:

)ssume is divisible b ' prove is divisible b .

Aivisible b b inductionAivisible b

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Proving a Property

%ase 0ase &n 8 5(:


)ssume is divisible b 6

is divisible b 6.

Aivisible b 5b case analsis

Aivisible b 6b induction

+rove

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Proving an $ne(uality

%ase 0ase &n 8 5(: is true


b induction

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)auchy-Sch!arz


+roof b induction &on n(: ?hen n81' L>S B8 >S.

?hen n85' want to show

0onsider

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)auchy-Sch!arz


Induction step: assume true for B8n' prove n71.

induction

b +&5(

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This Lecture




$ ) parado*

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Gray )oe

0an ou find an orderin# of all the nbit strin#s in such a wa thattwo consecutive nbit strin#s differed b onl one bitC

This is called the Dra code and has some applications.

>ow to construct themC Thin= inductivel@

5 bit

1111

bit

11111111111

1

0an ou see the patternC

>ow to construct 4bit #ra codeC

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Gray )oe

bit

111

111111111

bit &reversed(

111111

111111

111111111111111

111111111

11

111111

4 bit

differed b 1 bitb induction


differed b 1 bitb construction

,ver 4bit strin# appears e*actl once.

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Gray )oe

n bit

222

222212

n bit &reversed(

1222

22222

222222212

11

111111

n71 bit



differed b 1 bitb construction12

2

222222

So' b induction'

Dra code e*ists for an n.

,ver &n71(bit strin# appears e*actl once.

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Doal: tile the squares' e*cept one in the middle for %ill.

n2

n

2

Puzzle

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There are onl trominos &Lshaped tiles( coverin# three squares:

-or e*ample' for E * E pule mi#ht tile for %ill this wa:

Puzzle

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Theorem: -or an 5n * 5n pule' there is a tilin# with %ill in the middle.

+roof: &b induction on n (

+ &n ( ::8 can tile 5n * 5n with %ill in middle.

%ase case: &n 8(

&no tiles needed(

Puzzle

&Aid ou remember that we proved is divisble b C(

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n

2

Induction step: assume can tile 5n

* 5n

' prove can handle 5n7 1 * 5n7 1.

15 7n

Puzzle

FowwhatCC

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Puzzle

$ea* It would be nice if we could control the locations of the empt square.

n2

n2

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"one+

Puzzle

$ea* It would be nice if we could control the locations of the empt square.

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The ne! iea*

+rove that we can alwas find a tilin# with %ill any!here.

Puzzle

Theorem %: -or an 5n * 5n plaa' there is a tilin# with %ill anwhere.

Theorem: -or an 5n * 5n plaa' there is a tilin# with %ill in the middle.

0learl Theorem % implies Theorem.

) stron#er propert

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Theorem %: -or an 5n * 5n plaa' there is a tilin# with %ill anwhere.


+ &n ( ::8 can tile 5n * 5n with %ill anwhere.

%ase case: &n 8(

&no tiles needed(

Puzzle

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Induction step:

)ssume we can #et %ill anwhere in 5n * 5n .

+rove we can #et %ill anwhere in 5n 71 * 5n 71.

Puzzle

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Puzzle

Induction step:

)ssume we can #et %ill anwhere in 5n * 5n .

+rove we can #et %ill anwhere in 5n 71 * 5n 71.

n2

n2

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Method: Fow #roup the squares to#ether'

and fill the center with a tile.

"one+

Puzzle

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Some &emar,s

ote .: It ma help to choose a stron#er hpothesis

than the desired result &e.#. G%ill in anwhereH(.

?e need to prove a stron#er statement' but in

return we can assume a stron#er propert in

the induction step.

ote /: The induction proof of G%ill anwhereH implicitl

defines a recursive al#orithm for findin# such a tilin#.

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0aamar 1atri2

0an ou construct an n*n matri* with all entries 71 and

all the rows are ortho#onal to each otherC

Two rows are ortho#onal if their inner product is ero.

That is' let a 8 &a1' 2' an( and b 8 &b1' 2' bn('

their inner product ab 8 a1b1 7 a5b5 7 2 7 anbn

This matri* is famous and has applications in codin# theor.

To thin= inductivel' first we come up with small e*amples.

1 1

1 1

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0aamar 1atri2

Then we use the small e*amples to build lar#er e*amples.

Suppose we have an n*n >adamard matri* >n.

?e can use it to construct an 5n*5n >adamard matri* as follows.

>n >n

>n >n

So b induction there is a 5= * 5= >ardmard matri* for an =.

It is an e*ercise to chec= that the rows are ortho#onal to each other.

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$nuctive )onstruction

This technique is ver useful.

?e can use it to construct:

codes

#raphs

matrices

circuits

al#orithms

desi#ns

proofs

buildin#s

2

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This Lecture




$ ) parado*

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Parao2

Theorem: )ll horses have the same color.


Induction hpothesis:

+ &n ( ::8 an set of n horses have the same color

%ase case &n 8(:

Fo horses so obviousl true@

…

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&Inductive case(

)ssume an n horses have the same color.

+rove that an n7 1 horses have the same color.

Parao2

…n+1

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2

-irst set of n horses have the same color

Second set of n horses have the same color

&Inductive case(



Parao2

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2

Therefore the set of n 71 have the same color@

&Inductive case(



Parao2

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?hat is wron#C

+roof that + &n ( → + &n 71(

is false if n 8 1' because the two

horse #roups do not overlap .

-irst set of n8 1 horses

n 81

Second set of n8 1 horses

Parao2

&%ut proof wor=s for all n 3 1(

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4uic, Summary

ou should understand the principle of mathematical induction well'

and do basic induction proofs li=e

$ provin# equalit

$ provin# inequalit

$ provin# propert

Mathematical induction has a wide ran#e of applications in computer science.

In the ne*t lecture we will see more applications and more techniques.

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