Discrete Mathematics - NCKUmyweb.ncku.edu.tw/~hykao/course/dm/2009/slides/DMCH4.pdf · 自然數...

Discrete MathematicsCh t 4 P ti f th-- Chapter 4: Properties of the

Integers: Mathematical Induction

Hung-Yu Kao (高宏宇)Department of Computer Science and Information Engineering,

N l Ch K UNational Cheng Kung University

Outline

4.1 The Well-Ordering Principle: Mathematical InductionInduction4.2 Recursive Definitions

h i i i l i h i b4.3 The Division Algorithm: Prime Numbers4.4 The Greatest Common Divisor: The Euclidean Algorithm4 5 The fundamental Theorem of Arithmetic4.5 The fundamental Theorem of Arithmetic

Discrete Mathematics Discrete Mathematics –– CH4CH42009 Spring2009 Spring 2

自然數

自然數要適合五點(Peano axioms)：有一起始自然數 0。有一起始自然數 0任一自然數 a 必有後繼，記作 a +1。0並非任何自然數的後繼0 並非任何自然數的後繼。不同的自然數有不同的後繼。

（數學歸納公設）有一與自然數有關的命題設此（數學歸納公設）有一與自然數有關的命題。設此命題對 0 成立，而當對任一自然數成立時，則對其後繼亦成立則此命題對所有自然數皆成立後繼亦成立，則此命題對所有自然數皆成立。


4.1 The Well-Ordering Principle: g pMathematical Induction

The Well-Ordering Principle: Every nonempty subset of Z+contains a smallest element.( )}1|{}0|{ ≥∈=>∈=+ xxxx ZΖZ Z

+ is well ordered( )The Principle of Mathematical Induction: Let S(n) denote an open mathematical statement that involves one or more

f th i bl hi h t iti i t

}1|{}0|{ ≥∈>∈ xxxx ZΖZ

occurrences of the variable n, which represents a positive integer.a) If S(1) is true; and (basis step)b) If whenever S(k) is true, then S(k+1) is true. (inductive step)b) w e eve S(k) s ue, e S(k ) s ue. ( duc ve s ep)

then S(n) is true for all Using quantifiers

+∈Zn

)()]]]1()([[)([ 000 nSnnkSkSnknS ≥∀⇒+⇒≥∀∧


)()]]]()([[)([ 000

4.1 The Well-Ordering Principle: Mathematical Induction

P(n0)P(n0+1)P(n0+2)...


Why Induction WorksMore rigorously the validity of a proof by mathematical induction reliesMore rigorously, the validity of a proof by mathematical induction relies

on the well-ordering of Z+.Let S be a statement for which we have proven that S(1) holds and for p ( )

all n∈ Z+ we have S(n)⇒S(n+1). Claim: S(n) holds for all n∈ Z+

Proof by contradiction: Define the set F⊆ Z+ of values for which S does not hold: F = { m | S(m) does not hold}.

If F is non empt then F must have a smallest element ( ell ordering ofIf F is non-empty, then F must have a smallest element (well-ordering of Z+), let this number be z with ¬S(z). Because we know that S(1), it must hold that z>1. Because z is the smallest value, it must hold that S(z–1),

hi h di f f ll + ( ) ( )which contradicts our proof for all n∈ Z+: S(n)⇒S(n+1). Contradiction: F has to be empty: S holds for all Z+.


The Well-Ordering Principle: g pMathematical Induction

Ex 4.1 : For anyProof

2)1(

1 321,+

=+ =+⋅⋅⋅+++=∑∈ nnni nin Z

)1(

2)11(11

1

2)1(

1

1:)1( (i)

321 :)(Let +×

=

+=

==∑

=+⋅⋅⋅+++=∑

i

nnni

iS

ninS

12

)1(1

21

)1(321)1(Th(iii)

i.e., true,is )( Assume (ii)

)(( )

+

+=

=

∑

=∑

k

kkki

i

kkikS

ikS

2)1(1

1

11

)1()1()(

)1(321:)1(Then (iii)+

=

+=

++=++∑=

+++⋅⋅⋅+++=∑+kk

i

ki

kki

kkikS

2)2)(1( ++= kk

22321t +++ ++++∑ nnn iZDiscrete Mathematics Discrete Mathematics –– CH4CH42009 Spring2009 Spring 7

22

1321,try ++

=+ =+⋅⋅⋅+++=∈ ∑ nni nin Z


Ex 4.3 : Among the 900 three-digit integers (100 to 999), where the integer is the same whether it is read from left to right or from right to left are called palindromes Without actually determining all of theseleft, are called palindromes. Without actually determining all of these three-digit palindromes, we would like to determine their sum.

Solution

)10101(

1010110100 :palindrome typicalThe9 99 9∑ ∑ +∑ =

⎞⎜⎜⎛∑

+=++=

baaba

baabaaba

[ ]4510101010)101(10

)(

99 91 01 0

∑ ⋅+=∑ ⎥⎦

⎤⎢⎣

⎡∑+=

∑ ∑∑⎠

⎜⎜⎝∑

= == = a ba b

aba [ ]

500,4945091010

)(

911 0

=⋅+∑=

⎥⎦

⎢⎣ == = aa b

a


1=a


Ex 4.5For triangular number t =1+2+ +i= i(i+1)/2

22111 ⋅==t

2323

212⋅==

+=t

2436

3213⋅==

++=t

25410

43214⋅==

+++=t

ti=1+2+…+i= i(i+1)/2We want a formula for the sum of the first n triangular numbers.P fProof

][][)( )1(1)12)(1(11)1( 2 +=∑ +=∑∑ = ++++ iit nnnnnnn iin ][][)(

6)2)(1(

2262121 21

=

+=∑ +=∑∑ =

++===

iit

nnniii

i

Ex 4.4 prove it

700,171... 6)102)(101(100

6

10021 ==+++∴ ttt



Ex 4.5Consider pseudocode procedures (comparisons)P d 1 dditi d lti li ti ( dditi ll t i)Procedure 1: n additions and n multiplications (additionally, counter i)Procedure 2: 2 additions, 3 multiplications, and 1 division

procedure SumOfSquares1begin

procedure SumOfSquares2begin

( 1) (2 1)/6begin

sum:= 0for i:= 1 to n do

sum := sum + i2

sum:= n*(n+1)*(2*n+1)/6end

end



Ex 4.8 n 4n n2-7 n 4n n2-71 4 -62 8 -3

5 20 186 24 29).7(4,6For 2 −


nnHHH1

21

21 1, ,1 ,1 21 +⋅⋅⋅++=⋅⋅⋅+==

nHnH njn

−+=∑ )1(n, allFor

Ex 4.9 : Harmonic numbers

1)11(1:)1(Verify 111

1HHHS j

j−+===∑

=

jj=1Proof

)1(: true)( Assume

1

1kHkHkS kj

k k

k

j

j

−+=∑=

1)(k

])1([:)1(Verify 111

1 1

HkH

HkHkHHHkS kkkjjk

j

k

j

+−+=

+−+=+=+ ++∑ ∑+

= =

)1(2)(k

]111)[(k

1)(k

11

1

kH

HkkH

HkH

kk

kk

++=

+−+

−+=

++

++

+


. trueis )()1(2)(k 1

nSkH k

∴+−+= +


E 4 10Ex 4.10 :The elements of An are listed in ascending order, and |An|= 2n.To determine whether we must compare r with no more thanAr∈To determine whether , we must compare r with no more than n+1 elements in An.Proof

nAr∈

brccCbbBccbbCBccbbAS

aaaaAS==


M th ti l i d ti l j l i i ifi tiMathematical induction plays a major role in programming verification.Ex 4.11 :

The pseudocode program segment is supposed to produce the answer xyn.The pseudocode program segment is supposed to produce the answer xy .Proof

answer : true)( Assume

answer :)0(Verify 0

xykS

xyxSk=

==

findwenowagainloopwhile''theoftopthe return to and executed nsareinstructio loop the,first time for the

loop while'' theof top thereach the program the1,hen :)1(Verify knwkS +=+while n≠ 0 dobegin

x := x * y

1)1(k

findwenowagain,loop whiletheoftop the

1

knxyx

=−+=•=•

n := n - 1end

answer := x

true.is)()(answer final The

andwith continue willloopwhile''theAnd 1

1

,1

nSxyyxyyx

knyxkkk

∴===∴

=+


)(S


E 4 13Ex 4.13 :Show that for all we can express n using only 3’s and 8’s as summands (e g 14 = 3 + 3+ 8)

14≥nsummands. (e.g., 14 3 + 3+ 8)Proof

:true)(Assume kS

While :)1(Verify

:true)( Assume

=+

knkS

kS

1fors8'with twos3'fivereplace, s3' fiveleast at have sum the 14, sum, in the appears 8 no (ii) case

1for s3'ewith thre8replacesum,in theappears8oneleast at (i) case

+=∴≥

+=

kn

knQ

)1()(

1for s8with twos3fivereplace +⇒∴

+=kSkS

kn


Fibonacci SequenceConsider the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…Consider the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,…

which is defined by F1 = F2 = 1 and Fn+2 = Fn+1 + FnClearly this sequence F1,F2,… will grow, but how fast?C j t F ≥ (3/2)n f ll >10Conjecture: Fn ≥ (3/2)n for all n>10Proof by induction:

Base case: Indeed F11 = 89 ≥ (3/2)11 = 86.49756…

Inductive step for n>10:

Other base case: also F12 =144 ≥ (3/2)12 = 129.746338…Inductive step for n>10:

Assume Fn ≥ (3/2)n and Fn+1 ≥ (3/2)n+1, then indeedFn+2 = Fn + Fn+1 ≥ (3/2)n + (3/2)n+1 = (3/2)n(1 + (3/2))

2= (3/2)n(5/2) ≥ (3/2)n (9/4) = (3/2)n+2By induction on n, the conjecture holds.


Other Induction ProofsWe just saw a different kind of proof by induction where the inductive step is ∀n>10: [P(n), P(n+1) ⇒ P(n+2)]This time the basis step is proving P(11) and P(12).p p g ( ) ( )

There are many variations of proof by induction:Strong/ Complete induction: Here the inductive step is:Assume all of P(1),…,P(n), then prove P(n+1).The basis step is P(1) for this alternative form of induction.p ( )


Fibonacci NumbersThe sequence 1,1,2,3,5,8,13,21,34,… defined by Fn+2= Fn+Fn+1 is the famous Fibonacci sequence.

A closed expression of Fn isnn

511511 ⎞⎜⎛⎞⎜⎛ +n 2

5151

251

51F ⎟

⎠

⎞⎜⎜⎝

⎛ −−⎟

⎠

⎞⎜⎜⎝

⎛ +=

For large n, it grows like Fn ≈ 0.447214 × 1.61803n.This (1+√5)/2 ≈ 1.61803 is the Golden Ratio.


Fibonacci in NatureFibonacci numbers often occur in the natural world.Shape of shells: Number of petals

on flowers:

htt // h tt / j / /i d ht l


http://www.research.att.com/~njas/sequences/index.html


Theorem 4.2: The Principle of Mathematical Induction –Alternative Form: (or the Principle of Strong Mathematical I d ti )Induction)

Let S(n) denote an open mathematical statement that involves one or more occurrences of the variable n, which represents a positive p pinteger.a) If S(n0), S(n0+1), …, S(n1-1), and S(n1) are true

If h S( ) S( 1) S(k 1) d S(k) f

Basic step

b) If whenever S(n0), S(n0+1), …, S(k-1), and S(k) are true for some , then S(k+1) is also true

then S(n) is true for all1 where, nkk ≥∈

+Z

nn ≥ inductive stepthen S(n) is true for all .0nn ≥



E 4 14Ex 4.14 :Show that for all , n can be written as a sum of only 3’s and 8’s. (e g 14 = 3+3+8 15 = 3+3+3+3+3 16 = 8+8)

14≥n(e.g., 14 3+3+8, 15 3+3+3+3+3, 16 8+8)Proof

trueare)16(and)15()14(Verify SSS

:)1(Verify16 where true,are )( and )1(),2( , ),15(),14( Assume

true.are)16(and),15(),14(Verify

+≥−−⋅⋅⋅

kSkkSkSkSSS

SSS

trueis )1( trueis )2(,214 and ,3)2(1

:)1(Verify

+∴−≤−≤+−=+

+

kSkSkkkk

kSQ

)(

14 15 16 17 18 19 20 21 22 …



E 4 15Ex 4.15 :Show that

⎧

≤ where,3a nn

Proof⎩⎨⎧

≥∈++====

+−−− 3 where allfor ,

and,3,2,1

321

210

nnaaaaaaa

nnnn Z

11

000

332

331 (i)

=≤=

≤==

a

a

211

22

(ii)393

−−+ ++==≤=

kkkk aaaaa

1

21

3)3(3333 333

+

−−

==++≤

++≤kkkkk

kkk


Induction, another exampleTowers of Hanoi: (1883)Towers of Hanoi: (1883) We have three poles and n golden disksthe disks are only allowed in pyramid shape:the disks are only allowed in pyramid shape: no big disks on top of smaller onesHow to move the disks from one pole to another?How to move the disks from one pole to another?How many moves are requires?Call this number M(n)Call this number M(n).Note M(1)=1 and M(2)=3What about M(3)?W a abou (3)?


Inductive, Recursive

7 moves for 3 disks7 moves for 3 disksffIn general, 2In general, 2nn‐‐11


Making a ConjectureTo prove something by induction, you need a conjecture about the

general case M(n).

Here we have: M(1)=1, M(2)=3, M(3)=7,…Obvious conjecture… M(n) = 2n – 1 for all n > 0

Clearly, the basis step M(1)=1 holds.

Feeling for general n case: Each additional disk (almost) doubles the number of moves: M( +1) i t f t M( )M(n+1) consists of two M(n) cases…


Inductive StepHow to move n+1 disks, using an n-disk ‘subroutine’?

n nM(n) movesM(n) moves

n1 move

no e

M(n) moves

In sum: M(n+1) = 2 M(n)+1


( ) ( )

Proof of ToHWe just saw that for all n, we have M(n+1) = 2M(n)+1.

This enables our proof of the conjecture M(n)=2n–1.Proof: Basis step: M(1) = 21–1 = 1 holds.A th t M( ) 2n 1 h ld Th f t +1Assume that M(n)=2n–1 holds. Then, for next n+1:M(n+1) = 2 M(n) + 1 = 2(2n–1) + 1 = 2n+1 – 1.Hence it holds for n+1 End of proof by induction on nHence it holds for n+1. End of proof by induction on n.

With n=64 golden disks, and one move per second,this amounts to almost 600,000,000,000 years of work.


How to Prove InductivelyGeneral strategy:

1. Clarify on which variable you’re going to do the i d iinduction

2. Calculate some small cases n=1,2,3,…(Come up with your conjecture)your conjecture)

3. Make clear what the induction step n→n+1 is4 Prove the basis step prove the inductive step4. Prove the basis step, prove the inductive step,

and say that you proved it.


Induction in CSInductive proofs play an important role in computer science

because of their similarity with recursive algorithms.

Analyzing recursive algorithms often require the use of y g g qrecurrent equations, which require inductive proofs.


Structural InductionTh th d f i d ti l b li d t t tThe method of induction can also be applied to structures other than integers, like graphs, matrices, trees, sequences and so on. The crucial property that must hold is the well-ordered principle: there has to be a notion of size such that all objects have a finite size and each set of objects must have aobjects have a finite size, and each set of objects must have a smallest object.Examples: vertex size of graphs, dimension of matrices, depth of trees, length of sequences and so on.


L-Tiling an 2n×2n Square

Take a 2n×2n square with one tile missingCan you tile it with L-shapes?y p

Theorem: Yes, you can for all n.

Example for 8×8:


Example for 8×8:

Proving L-TilingBasis step of 2×2 squares is easy.

Inductive step: Assuming that 2n×2n tiling is possible,tile a 2n+1×2n+1 square by looking at 4 sub-squares:P t th 3 h l i th iddlPut the 3 holes in the middle:Put an L in the middle, and tile the sub squaresand tile the sub-squares using the assumption.

P f b i d tiProof by induction on n.This is a constructive proof.


4.2 Recursive DefinitionsE 4 19Ex 4.19 : Fibonacci numbers

1) F0 = 0; F1 = 1;2) F = F + F 2hllf ≥+Z2) Fn = Fn-1 + Fn-2, 2whereallfor ≥∈ + nn Z

10

2+

=

+ ×=∑∈∀ nnn

ii FFFn Z

Proof0=i

12

21

20

Assume

1*1 step, basis

+×=

=+

∑ kkk

i FFF

FF

21

21

2

10

Then

Assume

+

+

+=

+=

×

∑∑

∑

k

k

i

k

i

kki

i

FFF

FFF

211

00

)( ++

==

+×=+×=

∑∑

kkk

ii

FFFFFF


11

)(

++

++

×=+×=

kk

kkk

FFFFF

Recursive DefinitionsE 4 20Ex 4.20 : Lucas numbers

1) L0 = 2; L1 = 1;2) Ln = Ln-1 + Ln-2, 2 where allfor ≥∈ + nn Z

11 +−+ +=∈∀ nnn FFLn Z

312201 213,101 step, basis:Proof

+=+==+=+== FFLFFL

2111kk1k

11n

)()(Then 2k wherek,1,-k1,2,3,...,nfor Assume

−+−−+

+−

+++=+=≥=+=

kkkk

nn

FFFFLLLFFL

2

121

)()(

+

+−−

+=+++=

kk

kkkk

FFFFFF

Fn-2 Fn-1 Fn Fn+1 Fn+2

Ln-1 Ln Ln+1


1)1(1)1( ++−+ += kk FF

Recursive DefinitionsE 4 21 E l i bEx 4.21 : Eulerian numbers

0001,10 ,)1()( ,11,1,

kkmkakakma kmkmkm

4.3 The Division Algorithm: Prime gNumbers

D fi iti 4 1 If b di id d it b| if0d bZbDefinition 4.1: If we say b divides a, and write b|a, if there is an integer n such that a = bn, then b is divisor of a, or a is multiple of b.

,0and, ≠∈ bZba

We also say for b nonzero: b divides a or b is a factor of aTheorem 4.3:

,and,,integersthree theof twodividesand,Ife) allfor || d) |)]|()|[( c)

)]|()|[(b) 0)(a0|and|1a)

zyxazyxxbxabacacbba

baabbaaa

+=∈⇒⇒∧±=⇒∧≠Ζ

)(|theneachdividesIf1Forg)), ofn combinatiolinear called isy (y),(|)]|()|[( f)

integer. remaining thedivides then ,,,g,)

xcxcxcacacnicbcbxcbxacaba

ayy

+++∈≤≤++⇒∧

Ζ

Proof

)(|then ,each dividesIf.,1For g) 2211 nnii xcxcxcacacni +⋅⋅⋅++∈≤≤ Ζ

and | and | If f) ancambcaba ==⇒

y)(| )()()(y

cbxanymxayanxamcbx

+∴+=+=+∴


The Division Algorithm: Prime NumbersN b th i ti l li bl t l i d liNumber theory is now an essential applicable tool in dealing with computer and Internet security.

Primes are the positive integers that have only two positive divisors, namely, 1 and n itself. All other positive integers are called composite.


Sacks spiral, 1994Ulam spiral, 1963

The Division Algorithm: Prime NumbersL 4 1 If d i it th th i i+Lemma 4.1: If and n is composite, then there is a prime p such that p|n.

Proof

+∈ Zn

ProofIf no such a primeLet S be the set of all composite integers that have no prime divisors.By Well-Ordering Principle, S has a least element m.If m is composite, then m=m1m2, 1

The Division Algorithm: Prime NumbersTh 4 4 Th i fi it l i (E lid B k IX)Theorem 4.4: There are infinitely many primes. (Euclid, Book IX)

ProofIf notIf notLet p1 , p2, …, pk be the finite list of all primes.Let B = p1 p2 …pk+1.Let B p1 p2 …pk 1.Since B > pi , , B is not a prime.So B is composite, pj|B, . (Lemma 4.1)

ki ≤≤1kj ≤≤1j

Since pj|B and pj|p1 p2 …pk, so pj|1. (Contradiction, Theorem 4.3 (e))

integer.remainingthedividesthen, and ,, integers three the

of twodivides and , If e)

azyx

azyx +=


integer.remainingthedivides then a

The Division Algorithm: Prime NumbersTheorem 4.5: The Division Algorithm, if , with b > 0, then there exist Z∈ba ,unique

Proof

.0,with , brrqbarq /

=ab

rabWe call r the remainder when a is divided by b, and q the quotient when a is divided by b.

(1) q, r exist

φ≠∈=>>−∈−=

bbbbSSata

tbaZttbaS

)1()1(h1ld0If(ii), then ,0 and 0 If (i)}0,|{Let

0r,|(b) >/ ab

br ,| (c) −∴≤≤−

+−=−−=−−=≤Stbaab

bbabaatbaata,0 0 and 01

)1()1(then ,1let and0If(ii)Q

bSS P i i l )O d iW ll(0llhφ

SSbqacqbacbrbr

abbqabrqbarrSS

∈+−=⇒−=+=>/+==

−−=

The Division Algorithm: Prime Numbers

Ex 4.25If the dividend a =170 and the divisor b = 11, then the quotient q = 15 and the remainder r = 5 (170=15*11+5)15, and the remainder r = 5. (170=15*11+5)If the dividend a = 98 and the divisor b = 7, then the quotient q = 14, and the remainder r = 0. (98=14*7)f h di id d d h di i b h h iIf the dividend a = -45 and the divisor b = 8, then the quotient q = -6,

and the remainder r = 3. (-45=(-6)*8+3 or -45=(-5)*8-5?)Let +∈Zba,

If a = qb, then -a = (-q)b. So, the quotient is -q, and the remainder is 0.If a = qb + r, then -a = (-q)b - r = (-q - 1)b + (b - r). So, the quotient is -q -1, and the remainder is b - r.q ,



Ex 4.27 : Write 6137 in the octal system (base 8)Here we seek nonnegative integers r0 , r1 , r2, …, rk , 0< rk


Ex 4.29 : Two’s Complement Method: binary representation of negative integers

One’s Complement: interchange 0’s and 1’s.Add 1 to the prior result.

E l 610100001100101106 ≡→+→≡Example:Example: How do we perform the subtraction 33 – 15 in base 2 with patterns of 8-bits?

610100001100101106 −≡→+→≡

with patterns of 8 bits?

100010010(-15) 11110001(33) 00100001

+ 00001111 (+15)

General formula: x – y = x + [(2n – 1) – y + 1] – 2n

100010010


One’s complement of y

The Division Algorithm: Prime NumbersE 4 31 If d i i h h i i h+Ex 4.31 : If and n is composite, then there exists a prime p such that p|n and

Proof

+∈Zn.np ≤

Composite n = n1n2We claim that one of n1 , n2 must be less than or equal toIf not then d

,nIf not, then

tion)(contradic

and

21

21

nnnnnn

nnnn

=>=

>>

So, assume (i) If n1 is prime, the statement is true.

.1 nn ≤( ) 1 p ,(ii) If n1 is not prime, there exists a prime p < n1 where p|n1. (Lemma 4.1)

thenSo p|n and .np ≤

.1 nnp ≤<


p|

4.4 The Greatest Common Divisor: The Euclidean Algorithm

Definition 4.3: For , a positive integer c is called a greatest common divisor of (最大公因數gcd(a,b) )

Z∈ba,if ,ba

cdbadbacbcc|a

|havewe, andofdivisor common any for (b)), of divisorcommon a is ( | and (a)

QuestionsDoes a greatest common divisor of a and b always exist?

|,y( )

Does a greatest common divisor of a and b always exist?What would we deal with greatest common divisors for large integers aand b ?


The Greatest Common Divisor: The Euclidean Algorithm

Theorem 4.6: For all , there exists a unique that is the greatest common divisor of a and b.

f

+∈Zba,

}0|{l tGi bttbtSb + ZZ

+∈Zc

Proof

., ofdivisor common greast a is that claim We:Existence (i)Principle) Ordering-(Well element least a has

}0,,|{let ,,Given

baccS

btastsbtasSba >+∈+=∈ + ZZ

0,,| If |

4.3(f)) (Theorem then , and if ,,

crrqcaaccd

byd|axd|bd|abyaxcSc


The greatest common divisor of a and b is denoted by gcd(a, b).gcd(a, b) = gcd(b, a)Also aa =≠∈ 0)gcd(athen0aifZAlso,

gcd(a, b) = the smallest positive integer of the linear combination of a and b

d b ll d l ti l i h d( b) 1

aa ≠∈ 0)gcd(a,then ,0aif,Z

a and b are called relatively prime when gcd(a, b)=1i.e,

gcd(a, b)=c gcd(a/c, b/c)=1.1 with , =+∈ byaxyx Z

Ex 4.33gcd(42 70) = 14 42x+70y=14gcd(42, 70) = 14, 42x+70y=14

One solution x=2, y=-1x=2-5k, y=-1+3k



Theorem 4.7: Euclidean Algorithm: Let Set r0 = a and r1 = b and apply the division algorithm n times as follows:

., +∈Zbaand apply the division algorithm n times as follows:

rrrrqrrrrrqr


Proof

d)(|d|If , of divisor any for ,|Verify (i)

),gcd( at verify thTo

bbacrc

bar

n

n =

||and|Next)(e) 4.3 Theorem (|then

and),,(|and|If ,21102

21101010

rcrcrcrrqrrc

rrqrbrarrcrc

⇒+=

+===Qrrqr

rrqr+=+=

3221

2110

| and |Verify (ii)|down Continuing ||and|Next 321

brarrcrcrcrc

nn

n⇒⇒rrqr

+

+=⋅⋅⋅

4332

)(e) 4.3 Theorem ( equation last theFrom

,112 2|

1|

rrqrrrrr

nnnnnn

nn

+=∴ −−−−−

Qnnn

nnnn

rqrrrqr

=+=

−

−−−

1

112

)gcd(),(|[

]|[up,Continuing

100|]12|

1|23|

barbrarrrrrrr

rrrrrr

nnn

nnn

=∴==⇒∧

⇒∧


),gcd( barn =∴


Ex 4.34 : find the gcd(250,111), and express the results as a linear combination of these integers.

Solution


Ex 4.35 : the integers 8n+3 and 5n+2 are relatively prime.Solution

13120 )12()13(12525130 )13()25(138

+


An algorithm: a list of precise instructions designed to solve a particular type of problem, not just one special case.E 4 36 U E lid l ith t d l d (i d d )Ex 4.36 : Use Euclidean algorithm to develop a procedure (in pseudocode) that will find gcd(a, b) for all a, b ∈ Z+

procedure gcd(a,b: positive integers)p g ( , p g )begin

r:= a mod bd:= bwhile r > 0 do

c:= dd:= rr:= c mod d

endreturn(d)d { d( b) d}


end {gcd(a,b) = d}


Ex 4.37 : Griffin has two unmarked containers. One container holds 17 ounces and the other holds 55 ounces. Explain how Griffin can use his two containers to measure exactly one ouncecontainers to measure exactly one ounce.

Solution

1740 ,417355


Ex 4.38 : On the average, Brian debug a Java program in six minutes, but it takes 10 minutes to debug a C++ program. If he works for 104 minutes and doesn’t waste any time how many programs can be debug in eachand doesn t waste any time, how many programs can be debug in each language.

Solution =+⇒=+ 5253104106 yxyx

+≤≤+−+−=−⋅+⋅=⇒

−⋅+⋅=⇒=

352051040)352(5)5104(3)52(5104352

)1(52311)5,3gcd(

kykxkk

Q

⎧ ===

≤≤∴

+−=≤−=≤

214:18

3520,51040

5104

352

yxk

k

kykxQ

⎪⎩

⎪⎨

⎧

=========

⇒8,4:205 ,9:192,14:18

yxkyxkyxk


⎩


Theorem 4.8: If the Diophantine equation ax + by = c has an integer solution x = x0, y = y0 if and only if gcd(a, b) divides c. Definition 4 4: For c is called a common multiple of a b

,,, +∈Zcba

+∈ZcbaDefinition 4.4: For c is called a common multiple of a, bif c is a multiple of both a and b. Furthermore, c is the least common multiple of a, b if it is the smallest of all positive integers that are common

l i l f b W d b l ( b)

,,, ∈Zcba

multiples of a, b. We denote c by lcm(a, b).Theorem 4.9: Let with c = lcm(a, b). If d is a common multiple of a and b, then c|d.

,,, +∈Zcba, |

Proof

),(lcm0 ,| Ifmacbac

crrqcddc=∴=


Theorem 4.10: For ab = lcm(a, b).gcd(a, b)Ex 4.40 : By Theorem 4.10 we have

,, +∈Zba

(a) For all If a and b are relatively prime, then lcm(a, b) = ab.(b) The computations in Examples 4 36 (a = 168 b = 456) establish the

,, +∈Zba

(b) The computations in Examples 4.36 (a = 168, b = 456) establish the fact that gcd(168,458) = 24. As a result we find that lcm(a, b)?Solution

192,3)456,168(lcm

24)456,168gcd(

24456168 ==∴

=⋅

Q

24


4.5 The Fundamental Theorem of Arithmetic

Lemma 4.2: If and p is prime, then

.| || bpapabp ∨⇒,, +∈Zba

Proof||| ppp

|(ii)| (i)

apap/

)()(11),gcd(prime is

|( )

babypbxaypxapp

p

=+=+∴=∴Q

| and | )()(,1

abpppbabypbxaypx =+=+∴

Q y)(|)]|()|[( cbxacaba +⇒∧

Lemma 4.3:

(f)) 4.3(Theorem| bp∴

.1somefor || and prime is If .Let 21 niapaaappa ini ≤≤⇒∈ ⋅⋅⋅+Z


21 ini

The Fundamental Theorem of Arithmetic

Ex 4.41 : Show that is irrational.Proof

2

4 2)(Lemma|2|222

1),gcd( and ,, where,2 not If

2222 ⇒⇒=⇒=∴

=∈=⇒ +

aaab

baba

a

ba Z

i )( di)d(||4.2) (Lemma |2|2,2422

4.2)(Lemma |2|222

222222

2

⇒=⇒==⇒=∴

⇒⇒=⇒=∴

bbbbcbcabca

aaabb

tion)(Contradic2),gcd(|2|2 ≥∴∧ babaQ

p primeevery for irrational is p


The Fundamental Theorem of ArithmeticTheorem 4.11: Every integer n > 1 can be written as a product of primes y g p puniquely, up to the order of the primes. ( )

Proof

kksss pppn ⋅⋅⋅= 2211

existence(i)

mmmmmmmmm

m

)1(primes ofproduct as written becan , and,composite) is (

primes. ofproduct a as eexpressiblnot integer smallest theis not, If existence (i)

21 21

The Fundamental Theorem of ArithmeticEx 4.44 : How many positive divisors do 29,338,848,000 have? HowEx 4.44 : How many positive divisors do 29,338,848,000 have? How many of the positive divisors are multiples of 360? How many of the positive divisors are perfect squares?

Solution For(i) 21 pppn ksssSolution

117532000,848,338,29

)1()1)(1( is of divisors positive ofnumber The For (i)

335821

22

11

=

+⋅⋅⋅++⋅⋅⋅=

sssnpppn

k

kk

7651)1)(11)(311)(325)(138(532360 (ii)

17281)1)(11)(31)(35)(18(answer 23=

=+++++=∴

(0,2,4) choices 3 have we,5For )(0,2,4,6,8 choices 5 have we,8For (iii)

7651)1)(11)(311)(325)(138(answer

2

1

==

=+++−+−+−=∴

ss

6012235answer(0) choices 1 have we,1For

(0,2) choices 2 have we,3For

5

4,3

∴=

=s

ss


6012235answer =⋅⋅⋅⋅=∴

The Fundamental Theorem of Arithmetic

Ex 4.46 : Can we find three consecutive positive integers whose product is a perfect square, i.e., m(m+1)(m+2) = n2 , ?

Solution

+∈Znm,Solution

)4 4S tif21E i()2,1gcd(1)1,(gcdfact that the Use

exist ,such Suppose++==+ mmmm

nm

squareperfectaalsois)1(squareperfectais

| and ),2(|,| then ),1(| if , primeany For

)4.4Section of21Exercise(

2

2

+∴

+//+∴

mn

npmpmpmpp

Q

)1(12)2(

squareperfect a also is )2( squareperfect a alsois)1(squareperfect ais

222

+=++

Supplementary Exercise 34


Exercise (2009)

4-1 :18, 244-2 :14, 194 2 :14, 194-3 :4, 104-4 :10 144-4 :10, 144-5 :8


Discrete Mathematics - NCKUmyweb.ncku.edu.tw/~hykao/course/dm/2009/slides/DMCH4.pdf · 自然數...

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