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Discrete Mathematics

Relations and Functions

H. Turgut Uyar Aysegul Gencata Yayml Emre Harmanc

2001-2013


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c2001-2013 T. Uyar, A. Yayml, E. Harmanc

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Topics

1 RelationsIntroduction

Relation PropertiesEquivalence Relations

2 FunctionsIntroductionPigeonhole PrincipleRecursion


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Topics

1 RelationsIntroduction




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Relation

Definition

relation: A B C N

tuple: an element of a relation

A B: binary relation

ab is the same as (a, b)

representations:by drawingby matrix


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Relation

Definition

relation: A B C N






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Relation

Definition

relation: A B C N






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Relation

Definition

relation: A B C N






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Relation Example

Example

A = {a1, a2, a3, a4},B = {b1, b2, b3}

= {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)}

b1 b2 b3a1 1 0 1a2 0 1 1

a3 1 0 1a4 1 0 0

M =

1 0 10 1 1

1 0 11 0 0


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Relation Example

Example

A = {a1, a2, a3, a4},B = {b1, b2, b3}

= {(a1, b1), (a1, b3), (a2, b2), (a2, b3), (a3, b1), (a3, b3), (a4, b1)}

b1 b2 b3a1 1 0 1a2 0 1 1

a3 1 0 1a4 1 0 0

M =

1 0 10 1 1

1 0 11 0 0


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Relation Composition

Definition

relation composition:

let A B, B C= {(a, c) | a A, c C, b B [ab bc]}

M = M Musing logical operations:1 : T, 0 : F, : ,+ :


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Relation Composition Example

Example


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Relation Composition Matrix Example

Example

M =

1 0 00 0 10 1 10 1 0

1 0 1

M =1 1 0 00 0 1 10 1 1 0

M =

1 1 0 00 1 1 00 1 1 10 0 1 1

1 1 1 0


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Relation Composition Associativity

relation composition is associative

()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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()= ().

(a, d) ()

c [(a, c) (c, d) ]

c [b [(a, b) (b, c) ] (c, d) ]

b [(a, b) c [(b, c) (c, d) ]] b [(a, b) (b, d) ]

(a, d) ()


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Relation Composition Theorems

let , A B, andlet , B C

( ) =

( )

( )=

( )

( )


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)

C


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)

R l i C i i Th


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)

R l i C i i Th


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)

R l ti C iti Th


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( ) = .

(a, c) ( )

b [(a, b) (b, c) ( )]

b [(a, b) ((b, c) (b, c) )]

b [((a, b) (b, c) )

((a, b) (b, c) )]

(a, c) (a, c)

(a, c)

C R l ti


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Converse Relation

Definition

1 = {(b, a) | (a, b) }

M1 = MT

C s R l ti Th s


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Converse Relation Theorems

(1)1 =

( )1 = 1 1

( )1 = 1 1

1 = 1

( )1 = 1 1

1

1



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1 = 1.

(b, a) 1

(a, b)

(a, b) /

(b, a) / 1

(b, a) 1



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1 = 1.

(b, a) 1

(a, b)

(a, b) /

(b, a) / 1

(b, a) 1



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1 = 1.

(b, a) 1

(a, b)

(a, b) /

(b, a) / 1

(b, a) 1



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1 = 1.

(b, a) 1

(a, b)

(a, b) /

(b, a) / 1

(b, a) 1



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1 = 1.

(b, a) 1

(a, b)

(a, b) /

(b, a) / 1

(b, a) 1



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( )1 = 1 1.

(b, a) ( )1

(a, b) ( )

(a, b) (a, b)

(b, a) 1 (b, a) 1

(b, a) 1 1



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( )1 = 1 1.

(b, a) ( )1

(a, b) ( )

(a, b) (a, b)

(b, a) 1 (b, a) 1

(b, a) 1 1



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( )1 = 1 1.

(b, a) ( )1

(a, b) ( )

(a, b) (a, b)

(b, a) 1 (b, a) 1

(b, a)

1

1



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( )1 = 1 1.

(b, a) ( )1

(a, b) ( )

(a, b) (a, b)

(b, a) 1 (b, a) 1

(b, a)

1

1



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Co e se e at o eo e s

( )1 = 1 1.

(b, a) ( )1

(a, b) ( )

(a, b) (a, b)

(b, a) 1 (b, a) 1

(b, a)

1

1



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( )1 = 1 1.

( )1 = ( )1

= 1 1

= 1 1

= 1 1



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( )1 = 1 1.

( )1 = ( )1

= 1 1

= 1 1

= 1 1



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( )1 = 1 1.

( )1 = ( )1

= 1 1

= 1 1

= 1 1



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( )1 = 1 1.

( )1 = ( )1

= 1 1

= 1 1

= 1 1

Relation Composition Converse


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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11



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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11



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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11



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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11



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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11



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Theorem()1 = 11

Proof.

(c, a) ()1

(a, c)

b [(a, b) (b, c) ]

b [(b, a) 1 (c, b) 1]

(c, a) 11

Topics


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1 RelationsIntroductionRelation Properties

Equivalence Relations

2 FunctionsIntroduction

Pigeonhole PrincipleRecursion

Relation Properties


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A A

binary relation on A

let n mean

identity relation: E = {(x, x) | x A}

Relation Properties


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A A


let n mean


Relation Properties


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A A


let n mean


Reflexivity


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reflexive

A Aa [aa]

E

nonreflexive:a [(aa)]

irreflexive:a [(aa)]

Reflexivity


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reflexive

A Aa [aa]

E



Reflexivity


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reflexive

A Aa [aa]

E



Reflexivity


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reflexive

A Aa [aa]

E



Reflexivity Examples


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Example

R1 {1, 2} {1, 2}R1 = {(1, 1), (1, 2), (2, 2)}

R1 is reflexive

Example

R2 {1, 2, 3} {1, 2, 3}R2 = {(1, 1), (1, 2), (2, 2)}

R2 is nonreflexive



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Example

R1 {1, 2} {1, 2}R1 = {(1, 1), (1, 2), (2, 2)}

R1 is reflexive

Example

R2 {1, 2, 3} {1, 2, 3}R2 = {(1, 1), (1, 2), (2, 2)}

R2 is nonreflexive



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Example

R {1, 2, 3} {1, 2, 3}R = {(1, 2), (2, 1), (2, 3)}

R is irreflexive



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Example

R Z ZR = {(a, b) | ab 0}

R is reflexive

Symmetry


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symmetric

A Aa, b [(a = b) (ab ba) ((ab) (ba))]a, b [(a = b) (ab ba)]

1 =

asymmetric:a, b [(a = b) (ab (ba)) ((ab) ba))]

antisymmetric:

a, b [(a = b) (ab (ba))] a, b [(a = b) (ab) (ba)]

a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric


1 = asymmetric:a, b [(a = b) (ab (ba)) ((ab) ba))]

antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry


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symmetric



antisymmetric:


a, b [(ab ba) (a = b)]

a, b [(ab ba) (a = b)]

Symmetry Examples


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Example

R {1, 2, 3} {1, 2, 3}R = {(1, 2), (2, 1), (2, 3)}

R is asymmetric

Symmetry Examples


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Example

R Z ZR = {(a, b) | ab 0}

R is symmetric

Symmetry Examples


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Example

R {1, 2, 3} {1, 2, 3}R = {(1, 1), (2, 2)}

R is symmetric and antisymmetric

Transitivity


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transitive

A Aa, b, c [(ab bc) (ac)]

2

nontransitive:a, b, c [(ab bc) (ac)]

antitransitive:a, b, c [(ab bc) (ac)]

Transitivity


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transitive


2



Transitivity


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transitive


2



Transitivity


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transitive


2



Transitivity Examples


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Example

R {1, 2, 3} {1, 2, 3}R = {(1, 2), (2, 1), (2, 3)}

R is antitransitive

Transitivity Examples


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Example

R Z ZR = {(a, b) | ab 0}

R is nontransitive

Converse Relation Properties


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Theorem

The reflexivity, symmetry and transitivity properties

are preserved in the converse relation.

Closures


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reflexive closure:r = E

symmetric closure:s =

1

transitive closure:t = i=1,2,3,...

i = 2 3

Closures


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1


i = 2 3

Closures


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1


i = 2 3

Special Relations


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predecessor - successor

R Z ZR = {(a, b) | a b = 1}

irreflexive

antisymmetric

antitransitive

Special Relations


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adjacency

R Z ZR = {(a, b) | |a b| = 1}

irreflexive

symmetric

antitransitive

Special Relations


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strict order

R Z ZR = {(a, b) | a < b}

irreflexive

antisymmetric

transitive

Special Relations


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partial order

R Z ZR = {(a, b) | a b}

reflexive

antisymmetric

transitive

Special Relations


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preorder

R Z ZR = {(a, b) | |a| |b|}

reflexive

asymmetric

transitive

Special Relations


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limited difference

R Z Z,m Z+

R = {(a, b) | |a b| m}

reflexive

symmetric

nontransitive

Special Relations


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comparability

R U UR = {(a, b) | (a b) (b a)}

reflexive

symmetric

nontransitive

Special Relations


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sibling

irreflexive

symmetrictransitive

how can a relation be symmetric, transitive and nonreflexive?

Special Relations


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sibling

irreflexive

symmetrictransitive

how can a relation be symmetric, transitive and nonreflexive?

Topics


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1 RelationsIntroductionRelation PropertiesEquivalence Relations


Compatibility Relations


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Definition

compatibility relation:

reflexive

symmetric

when drawing, lines instead of arrows

matrix representation as a triangle matrix

1 is a compatibility relation



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Definition


reflexive

symmetric






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Definition


reflexive

symmetric




Compatibility Relation Example


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Example

A = {a1, a2, a3, a4}

R = {(a1, a1), (a2, a2),

(a3, a3), (a4, a4),

(a1, a2), (a2, a1),

(a2, a4), (a4, a2),

(a3, a4), (a4, a3)}

1 1 0 0

1 1 0 10 0 1 10 1 1 1

10 00 1 1



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Example

A = {a1, a2, a3, a4}

R = {(a1, a1), (a2, a2),

(a3, a3), (a4, a4),

(a1, a2), (a2, a1),

(a2, a4), (a4, a2),

(a3, a4), (a4, a3)}

1 1 0 0

1 1 0 10 0 1 10 1 1 1

10 00 1 1


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Example (1

)P: persons, L: languagesP = {p1, p2, p3, p4, p5, p6}L = {l1, l2, l3, l4, l5}

P L

M =

1 1 0 0 01 1 0 0 00 0 1 0 1

1 0 1 1 00 0 0 1 00 1 1 0 0

M1

=

1 1 0 1 0 01 1 0 0 0 1

0 0 1 1 0 10 0 0 1 1 00 0 1 0 0 0



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Example (1)

1 P P

M1 =

1 1 0 1 0 11 1 0 1 0 10 0 1 1 0 11 1 1 1 1 10 0 0 1 1 0

1 1 1 1 0 1

Compatibility Block


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Definition

compatibility block: C Aa, b [a C b C ab]

maximal compatibility block:not a subset of another compatibility block

an element can be a member of more than one MCB

complete cover: Cset of all MCBs

Compatibility Block


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Definition





Compatibility Block


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Definition





Compatibility Block Example


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Example (1)

C1 = {a4, a6}

C2 = {a2, a4, a6}

C3 = {a1, a2, a4, a6} (MCB)

C(A) = {{a1, a2, a4, a6},

{a3, a4, a6},{a4, a5}}



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Example (1)

C1 = {a4, a6}

C2 = {a2, a4, a6}

C3 = {a1, a2, a4, a6} (MCB)

C(A) = {{a1, a2, a4, a6},

{a3, a4, a6},{a4, a5}}



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Example (1)

C1 = {a4, a6}

C2 = {a2, a4, a6}

C3 = {a1, a2, a4, a6} (MCB)

C(A) = {{a1, a2, a4, a6},

{a3, a4, a6},{a4, a5}}



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Definition

equivalence relation:

reflexive

symmetric

transitive

equivalence classes (partitions)

every element is a member of exactly one equivalence class

complete cover: C



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Definition

equivalence relation:

reflexive

symmetric

transitive

equivalence classes (partitions)

every element is a member of exactly one equivalence class

complete cover: C

Equivalence Relation Example


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Example

R Z Z

R = {(a, b) | m Z [a b = 5m]}

R partitions Z into 5 equivalence classes

References


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Required Reading: Grimaldi

Chapter 5: Relations and Functions

5.1. Cartesian Products and Relations

Chapter 7: Relations: The Second Time Around

7.1. Relations Revisited: Properties of Relations7.4. Equivalence Relations and Partitions

Supplementary Reading: ODonnell, Hall, Page

Chapter 10: Relations

Topics


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2 FunctionsIntroductionPigeonhole Principle

Recursion

Functions


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Definition

function: f : X Yx X y1, y2 Y (x, y1), (x, y2) f y1 = y2

X: domain, Y: codomain (or range)

y = f(x) is the same as (x, y) f

y is the image of x under f

let f : X Y, and X1 Xsubset image: f(X1) = {f(x) | x X1}

Functions


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Definition






Functions


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Definition






Subset Image Examples


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Example

f : R R

f(x) = x2

f(Z) = {0, 1, 4, 9, 16, . . . }

f({2, 1}) = {1, 4}

Subset Image Examples


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Example

f : R R

f(x) = x2

f(Z) = {0, 1, 4, 9, 16, . . . }

f({2, 1}) = {1, 4}

Function Properties

Definition


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f : X Y is one-to-one (or injective):x1, x2 X f(x1) = f(x2) x1 = x2

Definition

f : X Y is onto (or surjective):y Y x X f(x) = y

f(X) = Y

Definition

f : X Y is bijective:f is one-to-one and onto

Function Properties

Definition


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Definition


f(X) = Y

Definition


Function Properties

Definition


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Definition


f(X) = Y

Definition


One-to-one Function Examples


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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0



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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0



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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0



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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0



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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0


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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0



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Example

f : R Rf(x) = 3x + 7

f(x1) = f(x2) 3x1 + 7 = 3x2 + 7 3x1 = 3x2

x1 = x2

Counterexample

g : Z Zg(x) = x4 x

g(0) = 04 0 = 0g(1) = 14 1 = 0

Onto Function Examples


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Example

f : R Rf(x) = x3

Counterexample

f : Z Zf(x) = 3x + 1

Onto Function Examples


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Example

f : R Rf(x) = x3

Counterexample

f : Z Zf(x) = 3x + 1

Function Composition


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Definition

let f : X Y, g : Y Z

g f : X Z

(g f)(x) = g(f(x))

function composition is not commutative

function composition is associative:f (g h) = (f g) h

Function Composition


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Definition

let f : X Y, g : Y Z

g f : X Z

(g f)(x) = g(f(x))

function composition is not commutative

function composition is associative:f (g h) = (f g) h

Function Composition Examples

Example (commutativity)


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f : R Rf(x) = x2

g : R R

g(x) = x + 5

g f : R R(g f)(x) = x2 + 5

f g : R R(f g)(x) = (x + 5)2




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f : R Rf(x) = x2

g : R R

g(x) = x + 5

g f : R R(g f)(x) = x2 + 5

f g : R R(f g)(x) = (x + 5)2




134/223


f : R Rf(x) = x2

g : R R

g(x) = x + 5

g f : R R(g f)(x) = x2 + 5

f g : R R(f g)(x) = (x + 5)2

Composite Function Theorems

Theorem


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Theoremlet f : X Y, g : Y Zf is one-to-one g is one-to-one g f is one-to-one

Proof.

(g f)(a1) = (g f)(a2) g(f(a1)) = g(f(a2)) f(a1) = f(a2)

a1 = a2


Theorem


136/223


Proof.


a1 = a2


Theorem


137/223


Proof.


a1 = a2


Theorem


138/223


Proof.


a1 = a2


Theorem


139/223

let f : X Y, g : Y Zf is one-to-one g is one-to-one g f is one-to-one

Proof.


a1 = a2


h


140/223

Theorem

let f : X Y, g : Y Zf is onto g is onto g f is onto

Proof.

z Z y Y g(y) = zy Y x X f(x) = y z Z x X g(f(x)) = z


Th


141/223

Theorem


Proof.



Th


142/223

Theorem


Proof.



Th


143/223

Theorem


Proof.


Identity Function


144/223

Definition

identity function: 1X

1X : X X1X(x) = x

Inverse Function


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Definition

f : X Y is invertible:

f1

: Y X [f1

f = 1X f f1

= 1Y]f1: inverse of function f

Inverse Function Examples

Example


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f : R Rf(x) = 2x + 5

f

1 : R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(x5

2 ) = 2x5

2 + 5 = (x 5) + 5 = x


Example


147/223

f : R Rf(x) = 2x + 5

f

1 : R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(x5

2 ) = 2x5

2 + 5 = (x 5) + 5 = x


Example


148/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(x5

2 ) = 2x5

2 + 5 = (x 5) + 5 = x


Example


149/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(x5

2 ) = 2x5

2 + 5 = (x 5) + 5 = x


Example


150/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(x5

2 ) = 2x5

2 + 5 = (x 5) + 5 = x


Example


151/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(

x5

2 ) = 2

x5

2 + 5 = (x 5) + 5 = x


Example


152/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(

x5

2 ) = 2

x5

2 + 5 = (x 5) + 5 = x


Example


153/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(

x5

2 ) = 2

x5

2 + 5 = (x 5) + 5 = x


Example


154/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(

x5

2 ) = 2

x5

2 + 5 = (x 5) + 5 = x


Example


155/223

f : R Rf(x) = 2x + 5

f1

: R Rf1(x) = x52

(f1 f)(x) = f1(f(x)) = f1(2x + 5) = (2x+5)52 =2x2 = x

(f f1

)(x) = f(f1

(x)) = f(

x5

2 ) = 2

x5

2 + 5 = (x 5) + 5 = x

Inverse Function

TheoremIf a function is invertible its inverse is unique


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If a function is invertible, its inverse is unique.

Proof.

let f : X Ylet g, h : Y X such that:g f = 1X f g = 1Yh f = 1X f h = 1Y

h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function

TheoremIf a function is invertible, its inverse is unique.


160/223


Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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, q

Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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, q

Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Inverse Function



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q

Proof.


h = h 1Y = h (f g) = (h f) g = 1X g = g

Invertible Function


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Theorem

A function is invertible if and only if it is one-to-one and onto.

Invertible Function

If invertible then one-to-one.

f A B

If invertible then onto.

f A B


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f : A B

f(a1) = f(a2)

f1(f(a1)) = f1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f : A B

f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

f : A B

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function


f : A B


f : A B


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f(a1) = f(a2)

f

1(f(a1)) = f

1(f(a2)) (f1 f)(a1) = (f

1 f)(a2)

1A(a1) = 1A(a2)

a1 = a2

b

= 1B(b)= (f f1)(b)

= f(f1(b))

Invertible Function

If bijective then invertible.f : A B


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f is onto b B a A f(a) = b

let g : B A be defined by a = g(b)

is it possible that g(b) = a1 = a2 = g(b) ?

this would mean: f(a1) = b = f(a2)

but f is one-to-one

Invertible Function



175/223





but f is one-to-one

Invertible Function



176/223





but f is one-to-one

Topics

1 RelationsIntroductionRelatio P o e ties


177/223



Pigeonhole Principle

Definition

Pigeonhole Principle (Dirichlet drawers):


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g ( )If m pigeons go into n holes and m > n,then at least one hole contains more than one pigeon.

let f : X Yif |X| > |Y| then f cannot be one-to-one

x1, x2 X [x1 = x2 f(x1) = f(x2)]

Pigeonhole Principle

Definition

Pigeonhole Principle (Dirichlet drawers):


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If m pigeons go into n holes and m > n,then at least one hole contains more than one pigeon.

let f : X Yif |X| > |Y| then f cannot be one-to-one

x1, x2 X [x1 = x2 f(x1) = f(x2)]

Pigeonhole Principle Examples

Example


180/223

Among 367 people, at least two have the same birthday.

In an exam where the grades integers between 0 and 100,how many students have to take the exam to make sure thatat least two students will have the same grade?

Pigeonhole Principle Examples

Example


181/223

Among 367 people, at least two have the same birthday.

In an exam where the grades integers between 0 and 100,how many students have to take the exam to make sure thatat least two students will have the same grade?

Generalized Pigeonhole Principle

Definition

Generalized Pigeonhole Principle:


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If m objects are distributed to n drawers,then at least one of the drawers contains m/n objects.

Example

Among 100 people, at least 9 (100/12) were bornin the same month.

Generalized Pigeonhole Principle

Definition

Generalized Pigeonhole Principle:


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If m objects are distributed to n drawers,then at least one of the drawers contains m/n objects.

Example

Among 100 people, at least 9 (100/12) were bornin the same month.

Pigeonhole Principle Example

Th


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Theorem

In any subset of cardinality 6 of the set S = {1,2,3,...,9},

there are two elements which total 10.


Theorem

Let S be a set of positive integers smaller than or equal to 14,

with cardinality 6. The sums of the elements

in all nonempty subsets of S cannot be all different.


185/223

Proof TrialA SsA : sum of the elements of A

holes:

1 sA 9 + + 14 = 69pigeons: 26 1 = 63

Proof.look at the subsets for which|A| 5

holes:

1 sA 10 + + 14 = 60pigeons: 26 2 = 62


Theorem





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holes:

1 sA 9 + + 14 = 69pigeons: 26 1 = 63


holes:

1 sA 10 + + 14 = 60pigeons: 26 2 = 62


Theorem





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holes:

1 sA

9 + + 14 = 69pigeons: 26 1 = 63


holes:

1 sA

10 + + 14 = 60pigeons: 26 2 = 62


Theorem





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holes:

1 sA

9 + + 14 = 69pigeons: 26 1 = 63


holes:

1 sA

10 + + 14 = 60pigeons: 26 2 = 62


Theorem





189/223


holes:

1 sA

9 + + 14 = 69pigeons: 26 1 = 63


holes:

1 sA

10 + + 14 = 60pigeons: 26 2 = 62


TheoremThere is at least one pair of elements among 101 elements

chosen from set S = {1, 2, 3, . . . , 200},th t f th l t f th i di id th th


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so that one of the elements of the pair divides the other.

Proof Method

we first show thatn !p [n = 2rp r N t Z [p = 2t + 1]]

then, by using this theorem we prove the main theorem


Theorem

There is at least one pair of elements among 101 elements

chosen from set S = {1, 2, 3, . . . , 200},th t f th l t f th i di id th th


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so that one of the elements of the pair divides the other.

Proof Method

we first show thatn !p [n = 2rp r N t Z [p = 2t + 1]]

then, by using this theorem we prove the main theorem


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence. Proof of uniqueness.


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n = 1: r = 0, p = 1

n k: assume n = 2

r

pn = k + 1:n = 2 : r = 1, p = 1n prime (n > 2) : r = 0, p = n(n prime) : n = n1n2

n = 2r

1 p1 2r

2 p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1

n k: assume n = 2

r


n = 2r

1 p1 2r

2 p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1

n k: assume n = 2

r


n = 2r

1 p1 2r

2 p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1n

k

: assumen

= 2

rp

n = k + 1:n = 2 : r = 1, p = 1n prime (n > 2) : r = 0, p = n(n prime) : n = n1n2

n = 2r

1 p1 2r

2 p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1n

k

: assumen

= 2

rp


n = 2r

1 p1 2r

2 p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1n

k: assume n = 2rp


n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]



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n = 1: r = 0, p = 1n k: assume n = 2rpn = k + 1:

n = 2 : r = 1, p = 1n prime (n > 2) : r = 0, p = n(n prime) : n = n1n2

n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence.

1 0 1

Proof of uniqueness.


199/223



n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence.

1 0 1



200/223



n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence.

1 0 1


f


201/223



n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence.

n 1 r 0 p 1


if i


202/223



n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem

n !p [n = 2rp r N t Z [p = 2t + 1]]

Proof of existence.

n 1: r 0 p 1


if i


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n = 2r1

p1 2r2

p2n = 2r1+r2 p1p2

if not unique:

n = 2r1 p1 = 2r2 p2

2r1r2 p1 = p2 2|p2


Theorem


chosen from set S = {1, 2, 3, . . . , 200}so that one of the elements of the pair divides the other.

Proof.


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T = {t | t S,i Z [t = 2i + 1]}, |T| = 100f : S T, r N olsuns = 2rt f(s) = t

if 101 elements are chosen from S, at least two of themwill have the same image in T: f(s1) = f(s2) 2

m1 t = 2m2 t

s1

s2=

2m1 t

2m2 t= 2m1m2


Theorem



Proof.


205/223



m1 t = 2m2 t

s1

s2=

2m1 t

2m2 t= 2m1m2


Theorem



Proof.


206/223



m1 t = 2m2 t

s1

s2=

2m1 t

2m2 t= 2m1m2


Theorem



Proof.


207/223

T= {

t|

t

S,

i Z [

t= 2

i+ 1]}, |

T| = 100

f : S T, r N olsuns = 2rt f(s) = t


m1 t = 2m2 t

s1

s2=

2m1 t

2m2 t= 2m1m2

Topics



208/223


Recursive Functions

Definitionrecursive function: a function defined in terms of itself

f(n) = h(f(m))


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inductively defined function: a recursive functionwhere the size is reduced at every step

f(n) =

k n = 0h(f(n 1)) n > 0

Recursion Examples

Example

f91(n) =

n 10 n > 100f91(f91(n + 11)) n 100


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Example (factorial)

f(n) =

1 n = 0n f(n 1) n > 0

Recursion Examples

Example

f91(n) =

n 10 n > 100f91(f91(n + 11)) n 100


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Example (factorial)

f(n) =

1 n = 0n f(n 1) n > 0

Euclid Algorithm

Example (greatest common divisor)

gcd(a, b) =

b b|agcd(b, a mod b) b a


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gcd(333, 84) = gcd(84, 333 mod 84)

= gcd(84, 81)

= gcd(81, 84 mod 81)

= gcd(81, 3)

= 3

Euclid Algorithm

Example (greatest common divisor)

gcd(a, b) =

b b|agcd(b, a mod b) b a


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gcd(333, 84) = gcd(84, 333 mod 84)

= gcd(84, 81)

= gcd(81, 84 mod 81)

= gcd(81, 3)

= 3

Fibonacci Series

Fibonacci series

Fn = fib(n) =

1 n = 11 n = 2fib(n 1) + fib(n 2) n > 2


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fib(n 1) + fib(n 2) n > 2

F1 F2 F3 F4 F5 F6 F7 F8 . . .1 1 2 3 5 8 13 21 . . .

Fibonacci Series

Theorem

ni=1 F

i2

= Fn

Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3


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n = k : ki=1

Fi2 = Fk Fk

+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2

=Fn

Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3


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n = k : ki=1

Fi2 = Fk Fk

+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2

=Fn

Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3


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n = k : ki=1

Fi2 = Fk Fk

+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2

=Fn

Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3

k


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n = k : ki=1

Fi2 = Fk Fk

+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2 = Fn Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3

k 2


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n = k : ki=1

Fi2 = Fk Fk

+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2 = Fn Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3

k 2


220/223

n = k : ki=1

Fi2 = Fk Fk+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Fibonacci Series

Theorem

ni=1

Fi2 = Fn Fn

+1

Proof.

n = 2 :2

i=1 Fi2 = F1

2 + F22 = 1 + 1 = 1 2 = F2 F3

k 2


221/223

n = k : ki=1

Fi2 = Fk Fk+1

n = k + 1 :

k+1i=1 Fi

2 =

k

i=1 Fi2 + Fk+1

2

= Fk Fk+1 + Fk+12

= Fk+1 (Fk + Fk+1)

= Fk+1 Fk+2

Ackermann Function

Ackermann function

ack(x y ) =

y + 1 x = 0ack(x 1 1) y = 0


222/223

ack(x, y) = ack(x 1, 1) y = 0ack(x 1, ack(x, y 1)) x > 0 y > 0

References

Required Reading: Grimaldi

Chapter 5: Relations and Functions

5.2. Functions: Plain and One-to-One5.3. Onto Functions: Stirling Numbers of the Second Kind5.5. The Pigeonhole Principle


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5.6. Function Composition and Inverse Functions

Supplementary Reading: ODonnell, Hall, Page

Chapter 11: Functions

04 relations functions 110317075641 phpapp02

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