functions and relationsphaas/info150/slides/s... · 2019. 10. 30. · functions and relations...

Functions and Relations

Reading: EC 4.1–4.5

Peter J. Haas

INFO 150Fall Semester 2019

Lecture 12 1/ 23

Functions and RelationsFunction Notation and TerminologyBinary RelationsInverse Relations and FunctionsComposition of FunctionsProperties of FunctionsOrdering RelationsEquivalence Relations

Lecture 12 2/ 23

Notation and Terminology of Functions 1

Definition

A function f : A ! B associates with each input from the domain A one and only oneoutput in the codomain B according to some rule

Terminology

I We say that “f is a function from A to B”

I If the rule associates to element a 2 A the element b 2 B, then we writef (a) = b and say that “f maps a to b” or “that value of f at a is b” or “f of aequals b”

Exercise: Define f : N ! N by the rule f (x) = 2x + 1

I Q: is every element of the codomain an output of one and only one input to f ?

Exercise: Define f : Z ! Z by the rule f (x) = x2


Lecture 12 3/ 23


Definition


Terminology

I We say that “f is a function from A to B”

I If the rule associates to element a 2 A the element b 2 B, then we writef (a) = b and say that “f maps a to b” or “that value of f at a is b” or “f of aequals b”

Exercise: Define f : N ! N by the rule f (x) = 2x + 1


Exercise: Define f : Z ! Z by the rule f (x) = x2


Lecture 12 3/ 23

No .

fix )=O,

i - e; 2×41=0

,

impliesX= - Igel R

NO . ft - 2) = Plz ) = 4-


Definition


Functions come in many guises

I Phone directories

I Word-processing software

I Addition: f (3, 4) = 7

I Truth tables: f : {T ,F}2 ! {T ,F}, e.g., f (p, q) = p ^ q

I Cutting the top card: (HCDS) = SHCD

Lecture 12 4/ 23

p q p ^ q

T T TT F FF T FF F F

Representing a Function

An example function

I Name: f

I Domain: {1, 2, 3, 4, 5}I Codomain: NI Rule: To each number in the domain, associate the square of the number

Representations of the rule

1. The above sentence

2. Algebraic formula: f (x) = x2

3. Set-based description: f = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25)}

4. Table:Input: 1 2 3 4 5

Output: 1 4 9 16 25

5. Arrow diagram:

Lecture 12 5/ 23

1 2 30

{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}

1 2 3 4 5

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250

Another functionexample:

r inputoutput

exactlyone outgoing →

arrow

Binary Relations

Definition

A binary relation consists of a domain A, a codomain B, and a subset of A⇥ B calledthe rule for the relation.

Example: Relation E

I Domain: The set S of all UMass students this semester

I Codomain: The set C of classes o↵ered at UMass this semester

I Rule: (x , y) is in E if student x is enrolled in class y this semester

Example: Relation L

I Domain: A = {1, 2, 3, 4}I Codomain: B = {2, 3, 5}I Rule: L = {(1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}I Succinct representation: L = {(x , y) 2 A⇥ B : x < y}I Infix notation: 1 L 2, 1 L 3, 2 L 5, 4 L 5, . . .

Observation

A function F : A ! B is a special case of a relation such that for every x 2 A, thereexists exactly one element y 2 B for which (x , y) 2 F (exactly one outgoing arrow)

Lecture 12 6/ 23

Relation L

2 3 5

1 2 3 4

ILL , 1<3,225,. .

.

Inverse Relations 1

Definition

Given a relation R with domain A and codomain B, the inverse R�1 of R is the

relation with domain B and codomain A such that

(x , y) 2 R if and only if (y , x) 2 R�1.

ExampleI Relation R: domain N and codomain Z with rule R = {(x , y) 2 N⇥ Z : x = y

2}or equivalently R = {(y2, y) : y 2 Z}

I Relation S : domain Z and codomain N with rule S = {(x , y) 2 Z⇥ N : y = x2}

or equivalently S = {(x , x2) : x 2 Z}

I Claim: R and S are inverses of each other1. If (x , y) 2 R, then x = y

2, which means that (y , x) = (y , y2) 2 S X2. If (x , y) 2 S , then x

2 = y , which means that (y , x) = (x2, x) 2 R X

Lecture 12 7/ 23

ex i ( 4,4 , ( 4,

-2 )

ex : 12,4 ),

C - 44 )

ed in verse of a L b C i.e.,

a ab ) from last slide

is G such that Ca,b) G G if a > b

ca,b)c. L iff a ab

,so ( b

,

a ) G G,

since b > a

Inverse Relations 2

Example: Relation E

I Domain is A = {1, 2, 3} and codomain is P(A)

I (x , y) 2 E (or equivalently x E y) if and only if x 2 y

I (y , x) 2 E�1 (or equivalently y E

�1x) if and only if x 2 y (also written y 3 x)

b ca

{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}b ca

{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}

Lecture 12 8/ 23

E E�1

" ycontains

x"

. I

Inverse Relations 3

Example: Arrow diagram when domain and codomain are the same

R = {(A, A), (A, B), (A, C), (A, E), (C , B), (C ,D), (E , A), (E , B), (E , C), (E ,D)}

R�1 = {(A, A), (B, A), (C , A), (E , A), (B, C), (D, C), (A, E), (B, E), (C , E), (D, E)}

A

B

D

E

C

A

B

D

E

C

Lecture 12 9/ 23

R R�1

Inverse Functions 1

Definition

Functions f : A ! B and g : B ! A are inverses of each other if

f (a) = b if and only if g(b) = a

for all a 2 A and b 2 B. We often write f�1 for the inverse of f .

Example: Prove that f : Z ! Z with rule f (x) = x + 3 and g : Z ! Z with ruleg(y) = y � 3 are inverses of each other

I Claim 1: For all a 2 A and b 2 B, if f (a) = b then g(b) = a

I Let a, b 2 Z be given such that f (a) = b, i.e., a+ 3 = b

I Then a = b � 3, i.e., g(b) = a. X

I Claim 2: For all a 2 A and b 2 B, if g(b) = a then f (a) = b

I Let a, b 2 Z be given such that g(b) = a, i.e., b � 3 = a

I Then a+ 3 = b, i.e., f (a) = b. X

Lecture 12 10/ 23

Inverse Functions 1

Definition

Functions f : A ! B and g : B ! A are inverses of each other if

f (a) = b if and only if g(b) = a

for all a 2 A and b 2 B. We often write f�1 for the inverse of f .

Example: Prove that f : Z ! Z with rule f (x) = x + 3 and g : Z ! Z with ruleg(y) = y � 3 are inverses of each other

I Claim 1: For all a 2 A and b 2 B, if f (a) = b then g(b) = a

I Let a, b 2 Z be given such that f (a) = b, i.e., a+ 3 = b

I Then a = b � 3, i.e., g(b) = a. X

I Claim 2: For all a 2 A and b 2 B, if g(b) = a then f (a) = b

I Let a, b 2 Z be given such that g(b) = a, i.e., b � 3 = a

I Then a+ 3 = b, i.e., f (a) = b. X

Lecture 12 10/ 23

-

→

←

Inverse Functions 2: Computing an Inverse

Example: for f : Q ! Q with rule f (x) = 25 x � 2, find f

�1

I Let a, b 2 Q be given such that f (a) = b, i.e., 25a� 2 = b

I Solving for a, we have a = 52b + 5

I So take f�1(y) = g(y) = 5

2 y + 5

Lecture 12 11/ 23

Inverses and Arrow Diagrams

An inverse is obtained by reversing the arrows

2

3

4

a

b

c

d

1

b

c

d

1

2

3

4

a

Example: Why is there is no function whose inverse g : {a, b, c, d} ! {1, 2, 3, 4} isgiven below?

2

3

4

a

b

c

d

1

b

c

d

1

2

3

4

a

Lecture 12 12/ 23

g-' ?

g. yrs ) = ?

g-'

air ?

Composition of Functions

Definition

Given f : A ! B and g : B ! C , the composition g � f of g and f has domain A,codomain C and rule (g � f )(x) = g(f (x)).

Example:

I f : R�0 ! R with rule f (x) =px

I g : R ! R with rule g(x) = 2x

I Then (g � f )(x) = g(f (x)) = g(px) = 2

px

I Then (f � g)(x) = f (g(x)) = f (2x) =p2x (what is the problem here?)

Composition via arrow diagrams

Lecture 12 13/ 23

a

b

c

d

x

y

z

a

b

c

d

2

1

2

1x

y

z

a

b

c

d2

1x

y

z

yes::b.

ate

,evaluate

first

domain of fog = domain of g= IR

,but - I ER and

I fog)th -

- Fn 4 IR

if we define 5 : Pio → 11230 with rule gun-

- H

Ithen fog is well defined

Inverse Functions Revisited

Definition

For a given set A, the identity function on A is the function ◆A : A ! A with the rule◆A(x) = x for all x 2 A. We’ll often simply write ◆ when A is clear from context. Wecan also write ◆A = {(x , x) : x 2 A} when we wish to view ◆A as a binary relation.

Theorem

Functions f : A ! B and g : B ! A are inverses of each other if and only iff � g = ◆B and g � f = ◆A. In other words, f �1

�f (x)

�= x and f

�f�1(y)

�= y

Example 1:I Let f : Q ! Q be the function with rule f (x) = 2

5 x � 2

I Let g : Q ! Q be the function with rule g(x) = 52 x + 5

I Then (g � f )(x) = g(f (x)) = g( 25 x � 2) = 52 (

25 x � 2) + 5 = (x � 5) + 5 = x

I Also, (f � g)(x) = f (g(x)) = f ( 52 x + 5) = 25 (

52 x + 5)� 2 = (x + 2)� 2 = x

Example 2: f : A ! A⇥ A with f (a) = (a, a) and g : A⇥ A ! A with g(x , y) = x

I Given a 2 A: (g � f )(a) = g(f (a)) = g(a, a) = a, so g � f = ◆AI Given (1, 2) 2 A⇥ A: (f � g)(1, 2) = f (1) = (1, 1) 6= (1, 2), so f � g 6= ◆A⇥A

Lecture 12 14/ 23

Inverse Functions Revisited

Definition

For a given set A, the identity function on A is the function ◆A : A ! A with the rule◆A(x) = x for all x 2 A. We’ll often simply write ◆ when A is clear from context. Wecan also write ◆A = {(x , x) : x 2 A} when we wish to view ◆A as a binary relation.

Theorem

Functions f : A ! B and g : B ! A are inverses of each other if and only iff � g = ◆B and g � f = ◆A. In other words, f �1

�f (x)

�= x and f

�f�1(y)

�= y

Example 1:I Let f : Q ! Q be the function with rule f (x) = 2

5 x � 2

I Let g : Q ! Q be the function with rule g(x) = 52 x + 5

I Then (g � f )(x) = g(f (x)) = g( 25 x � 2) = 52 (

25 x � 2) + 5 = (x � 5) + 5 = x

I Also, (f � g)(x) = f (g(x)) = f ( 52 x + 5) = 25 (

52 x + 5)� 2 = (x + 2)� 2 = x

Example 2: f : A ! A⇥ A with f (a) = (a, a) and g : A⇥ A ! A with g(x , y) = x

I Given a 2 A: (g � f )(a) = g(f (a)) = g(a, a) = a, so g � f = ◆AI Given (1, 2) 2 A⇥ A: (f � g)(1, 2) = f (1) = (1, 1) 6= (1, 2), so f � g 6= ◆A⇥A

Lecture 12 14/ 23

fawnby

notinverses

Properties of Functions 1

Definition

The function f : A ! B is invertible if there is a function f�1 : B ! A such that

f (x) = y if and only if f �1(y) = x . By symmetry of the definition, (f �1)�1 = f .

Example

I With A = B = R�0, if f has rule f (x) = x2, then f

�1(x) =px

I Arrow diagram:

A non-invertible function g

I Problem 1: no arrow points to 4 (g is not onto)

I Problem 2: two arrows point to 3 (g is not one-to-one)

Lecture 12 15/ 23

2

3

4

a

b

c

d

1

b

c

d

1

2

3

4

a

(f �1�f )(x) = x


Definition

The function f : A ! B is invertible if there is a function f�1 : B ! A such that

f (x) = y if and only if f �1(y) = x . By symmetry of the definition, (f �1)�1 = f .

Example

I With A = B = R�0, if f has rule f (x) = x2, then f

�1(x) =px

I Arrow diagram:

A non-invertible function g

I Problem 1: no arrow points to 4 (g is not onto)

I Problem 2: two arrows point to 3 (g is not one-to-one)

Lecture 12 15/ 23

2

3

4

a

b

c

d

1

b

c

d

1

2

3

4

a

2

3

4

a

b

c

d

1

b

c

d

1

2

3

4

a

(f �1�f )(x) = x

-


Definition

The function f : A ! B is onto if 8y 2 B, 9x 2 A, f (x) = y . The function f isone-to-one if (x 6= y) !

�f (x) 6= f (y)

�.

Theorem

A function f : A ! B is invertible if and only if it is both one-to-one and onto.

Exercise: Which functions are invertible?

I f : Z ! Z with f (x) = 2x + 3

I g : Z ! N with g(x) =

(�2z if z 0

2z � 1 if z > 0

I h : N ! N with h(n) = sum of digits in the numeral n

Lecture 12 16/ 23

2

3

4

a

b

c

d

1

X

not onto.

flex ) -

-O implies LXt3=o ,

so HI -3g ¢ #✓ gt - 11--2

, go-25-4,91-31--6, - -

-

c. to . I and onto 90=9 guk '

9123=3,433--5,

. --

,

"

Anot I - tot : htt 3) = honor ) -

- 4

(Direct) Proofs About Functions

Proposition 1

If f : A ! B is one-to-one and g : B ! C is one-to-one, then (g � f ) : A ! C isone-to-one.

Proof

1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)

2. This means that g(f (x1)) = g(f (x2))

3. Since g is one-to-one, this means that f (x1) = f (x2)

4. Since f is one-to-one, this means that x1 = x2 X

Example: f : N ! N wih f (x) = 5x + 7 and g : N ! Q with g(n) = 5n+2


2. This means that g(f (x1)) = g(f (x2)) or 5f (x1)+2 = 5

f (x2)+2 or f (x1)+25 = f (x2)+2

5

3. Multiply by 5 and subtract 2 on both sides: f (x1) = f (x2) or 5x1 + 7 = 5x2 + 7

4. Subtract 7 and divide by 5 on both sides to get x1 = x2 X

Lecture 12 17/ 23

I - to -I : ( x

, + 4) → Lf exit ftp.D

Xcx,k f Hr ) ) → cry ,

= XD


Proposition 1


Proof








f (x2)+2 or f (x1)+25 = f (x2)+2

5



Lecture 12 17/ 23

of contrapositive I flexi -

-flair ) ) → µ,- uhh )


Proposition 1


Proof








f (x2)+2 or f (x1)+25 = f (x2)+2

5



Lecture 12 17/ 23

,Tracing the proof

Properties of Relations

Definition

Let R be a binary relation on a set A

1. R is reflexive if (a, a) 2 R for all a 2 A (must have loops)

2. R is antisymmetric if (a, b) 2 R and a 6= b implies (b, a) 62 R (no double arrows)

3. R is transitive if (a, b), (b, c) 2 R implies (a, c) 2 R (if you can get from a to c

following two arrows, you can also get there following one arrow)

Definition

A relation R on a set A is partial order if it is antisymmetric, transitive, and reflexive

Examples

1. A = {1, 2, 3, 4}: a R1 b means a b

2. A = P({1, 2, 3}): a R2 b means a ✓ b

3. A = {1, 2, 3, 6}: a R3 b means a divides b

Lecture 12 18/ 23

L-

- { C 1,27 ,( 3,4)

, ④7) I 122,324 ,"

LE s fl :D ,134 ,

.. . I

)


Definition






Definition


Examples

1. A = {1, 2, 3, 4}: a R1 b means a b

2. A = P({1, 2, 3}): a R2 b means a ✓ b


Lecture 12 18/ 23


Definition






Definition


Examples

1. A = {1, 2, 3, 4}: a R1 b means a b

2. A = P({1, 2, 3}): a R2 b means a ✓ b


Lecture 12 18/ 23

3

2

1

1

2 3

6

1

4{1,2,3}

{1,3}{2,3}

{1}{2}

{3}

{}

{1,2}

R1 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)} Hasse diagrams=p

antisymmetric : ( atb )s7[ca ,HeRrcb,aIER ]

Proofs About Properties

Example: Prove the reflexive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let a 2 Z be given

2. Since a� a = 0, which is even, we have that (a, a) 2 R X

Example: Prove the transitive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let (a, b), (b, c) 2 R be given [We’ll prove that (a, c) 2 R]

2. Since a� b and b � c are even, we can write a� b = 2K and b � c = 2L forsome integers K and L

3. Therefore a� c = (a� b) + (b � c) = 2K + 2L = 2(K + L)

4. Thus (a� c) is even and hence (a, c) 2 R X

Example: Prove the antisymmetric property for R = {(s, t) 2 P({1, 2, 3, 4})2 : s ✓ t}I Show: for all s, t 2 P({1, 2, 3, 4}), if (s, t) 2 R and (t, s) 2 R, then s = t

1. Let s, t 2 P({1, 2, 3, 4}) be given2. Since (s, t) 2 R and (t, s) 2 R, we have that s ✓ t and t ✓ s

3. It follows that s = t by definition of set equality X

Lecture 12 19/ 23

Proofs About Properties

Example: Prove the reflexive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let a 2 Z be given

2. Since a� a = 0, which is even, we have that (a, a) 2 R X

Example: Prove the transitive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let (a, b), (b, c) 2 R be given [We’ll prove that (a, c) 2 R]

2. Since a� b and b � c are even, we can write a� b = 2K and b � c = 2L forsome integers K and L

3. Therefore a� c = (a� b) + (b � c) = 2K + 2L = 2(K + L)

4. Thus (a� c) is even and hence (a, c) 2 R X

Example: Prove the antisymmetric property for R = {(s, t) 2 P({1, 2, 3, 4})2 : s ✓ t}I Show: for all s, t 2 P({1, 2, 3, 4}), if (s, t) 2 R and (t, s) 2 R, then s = t

1. Let s, t 2 P({1, 2, 3, 4}) be given2. Since (s, t) 2 R and (t, s) 2 R, we have that s ✓ t and t ✓ s

3. It follows that s = t by definition of set equality X

Lecture 12 19/ 23

( at b) → Tla,

b) GR n lb,at ER ]

contrapositive : ca , b) f R A lb, a) ER# (a =D

Other Types of Orders

Definition

A relation R over a set A is irreflexive if (a, a) 62 R for all a 2 A. A strict partialordering on A is a relation R on A that is transitive, antisymmetric, and irreflexive.

Notes

I Irreflexive means no loops in an arrow diagram

I A relation R can be neither reflexive or irreflexive if some (but not all) nodes inthe arrow diagram have loops

Example:

I Strict subset relation: write A ⇢ B if A ✓ B and B � A 6= ;I Then R = {(A,B) 2 P({1, 2, 3, 4})2 : A ⇢ B} is a strict partial ordering

Definition

A relation R on A is a total ordering if it is a partial ordering and also satisfies theproperty:For all a, b 2 A, if a 6= b, then either (a, b) 2 R or (b, a) 2 R. A strict total orderinghas the same properties except that it is irreflexive.

Lecture 12 20/ 23

c :p no Boo

-

i.e. ,B has at least one element x

set . x ¢ A

Types of Orderings: Examples

Notation:

I A = {1, 2, 4, 8}, B = P({1, 2, 3}), C = {0, 1}4

I V (↵) = value of binary numeral ↵, e.g., V (0101) = 5

Example 1: R1 = {(x , y) 2 A2 : x y} total ordering

Example 2: R2 = {(x , y) 2 B2 : x ✓ y} partial ordering (incomparable subsets)

Example 3: R2 = {(S ,T ) 2 B2 : every element in S is every element in T} ?

Example 4: R2 = {(S ,T ) 2 B2 : n(S) < n(T )} ?

Example 5: R2 = {(S ,T ) 2 B2 : sum of elements in S is sum of elements in T} ?

Example 6: R2 = {(↵,�) 2 C2 : ↵ has fewer 1’s than � has} ?

Example 7: R2 = {(↵,�) 2 C2 : V (↵) V (�)} ?

Lecture 12 21/ 23

Base 10 : 234=4×1001-3×10 't 2140LBase 2 : 101 I l XI to . 2 't 1.22

= s

e.9 .1011

,

1100 ,-

--

+= go.info,

13×40,13×10,13

focuson

①Reflexive

lirncfkx.ve.

non. strict

I strict

② can youcompare

any pairof

element

11,449433and 1531411,23 yess

both

no ,partial

}partial ordering : reflexive

, lip ) huh incomparable orderingstrict partial ordering4 irreflexivc , 41,4 ,

$3,43,

incomparable

§ a

Total ordering : antisymmetric ,

transitive, reflexive ,

not incomparable

•strict partial order . 1100 and 0011 incomparable

7Total ordering cis condition were : Vlade Up)

,then strict total ordering

Equivalence Relations

Definition

A partition of a set A is a set S = {S1, S2, S3, . . .} such that

1. For all i , Si 6= 0

2. For all i , j : if Si 6= Sj , then Si \ Sj = ;3. S1 [ S2 [ S3 [ · · · = A

Definition

A relation R on A is an equivalence relation if there is a partition S of A such that(x , y) 2 R if and only if x and y are in the same part of S. We call S the partition ofA induced by R.

Example 1: A = {1, 2, 3, 4, 5, 6}I R = {(a, b) 2 A⇥ A : a� b is even}I S = {{1, 3, 5}, {2, 4, 6}}

Example 2:

I R = {(a, b) 2 Z⇥ Z : a� b is divisible by 4}I S = {P0,P1,P2,P3} where Pi = {a 2 Z : a = 4k + i for some k 2 Z}

Lecture 12 22/ 23

disjoint


Definition




Definition



Example 2:


Lecture 12 22/ 23

disjoint

part ion induced by R

odd - Odd is even

even- even is even

even- odd is even ,

odd - even is even


Definition




Definition



Example 2:


Lecture 12 22/ 23

if a,be Pi

,

then a- 4kt is bintifor integers k and L

So a- b -

- 14kt i ) - Ltthti ) -

- 41k - L ),which is div . by 4

-

✓

Formal Properties of an Equivalence Relation

Definition

A relation R on A is symmetric if for all a, b 2 A, if (a, b) 2 A then (b, a) 2 A.

Theorem

A relation R on A is an equivalence relation if and only if it is reflexive, symmetric,and transitive.

Example: For each relation, determine whether it is an equivalence relation

I T1 = {(a, b) 2 Z2 : b � a is divisible by 5}

I T2 = {(a, b) 2 Z2 : a2 � b2 is divisible by 5}

I T3 = {(a, b) 2 Z2 : |a� b| 2}

Lecture 12 23/ 23

[email protected],

thenb-saymm-elgnj.bydub 'S

YGsimilar ③ a - b - 5k b - Ct 5h

a - Coca- b) tcb - c) ask -5L

NO V - 51k - D4" GT3 ↳ 4) ET,

Transitive

( 54 ) ¢T ,not transitive ie . II Land 2%4 but not 1154

functions and relationsphaas/info150/slides/s... · 2019. 10. 30. · functions and relations...

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