1 Annoucement n Skills you need: (1) (In Thinking) You think and move by Logic Definitions...

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Annoucement

Skills you need:

(1) (In Thinking) You think and move by• Logic• Definitions• Mathematical properties (Basic algebra etc.)

(2) (In Exploration) You understand by:• Trying examples• Drawing diagrams

– For relations, 3 visualization tools have been given to you. Be flexible in using your alternatives.

(3) Using (1) and (2) to understand new things.

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Lecture 3 (part 1)

FunctionsReading: Epp Chp 7.1, 7.3, 7.5

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Overview

Functions– Definition– Visualization Tool– Examples– Notation– Identity Function

Definitions– Domain, codomain,

range– Image, inverse image– Let f:A B and let S A

and T B.• S InvImgf(Imgf(S))

• Imgf(InvImgf(T)) T

Operations on Functions– Set related (,,):

1. If fg is a function, then f = g2. If fg is a function, then f = g3. If fg is a function, then fg=

– Inverse– Composition– Theorems

1. G o f is a function.

2. F o idA idB o f f3. (h o g) o f = h o (g o f)4. (g o f)-1 = f-1 o g-1

Special Kinds of Functions– Injective, Surjective,

Bijective– Theorems

• Preservation under ‘o’.• On inverses

Multi-argument functions

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1. Definition: Function

Given a relation R from a set A to a set B, R is a function iff1. xA, yB, x R y

2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

A function is a relation. It’s just a special kind of relation – a relation with 2 restrictions.

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

1. Every element in A must be associated with AT LEAST one element in B.

Extract the meaning from the logical expression.

A BR

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

2. An element in A can only be associated with AT MOST one element in B.


A BR

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

1 and 2 taken together: Every element in A can only be associated with EXACTLY ONE element in B.


A BR

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2. Visualisation Tool: Arrow Diagram


2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

A Bf

Not a function

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

Yes, it’s a function.

A Bf

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2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

Not a function

A Bf

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3. Examples


2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

Example 1:

Is R a function?

No. (Cond 1)

No. (Cond 2)

Yes.

R = {(1,2)}

R A x BA

{1,2,3}

B

{1,2,3}

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

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

{1,2,3,4} {1,2,3} R = {(1,1),(2,2),(3,3)} No. (Cond 1)

{1,2,3} {1,2,3,4} R = {(1,1),(2,2),(3,3)} Yes.

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3. Examples


2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

Example 2: Let R Q x Z such that…

(i) x R y iff x = y.

Q: Is R a function?

A: No (1st Condition: ½ maps to nothing)

Q Z

½ ?

(ii) (a/b) R c iff a.b = c.

Q: Is R a function?

A: Yes

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3. Examples


2. xA, y1,y2B, x R y1 x R y2 y1 = y2.

Example 3: Let R Z x Z such that…

(i) x R y iff y = x2.

Q: Is R a function?

A: Yes.

(ii) x R y iff x = y2.

Q: Is R a function?

A: No. (1st and 2nd Condition violated)

Z Z

3 ?

1 1

-1

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3. Examples

Functions in real life:1. Hamming distance function (p351).

2. Encoding/decoding functions (p351).

3. Boolean functions (p352).

4. A program is a function.

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Eg. 3: F Z x {0,1}, a F b iff (a is even b=1) (a is odd b=0)

4. Notation

We usually use “f,g,h,F,G,H” to denote functions. If the relation f A x B is a function, we write it as:

f : A B If there is a way to compute yB from any given

xA, we usually write ‘f(x)’ in place of ‘y’.

We will write it as:Eg. 1: F Z x Z , x F y iff y = x2. ‘F’ is a function.

F : Z Z, F(x) = x2

We will write it as:Eg. 2: F Z x Z , x F y iff y = x2 + 2x + 1

F : Z Z, F(x) = x2 + 2x + 1

F : Z {0,1}, F(x) =

1, if x is even

0, otherwise

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4. Notation

We usually use “f,g,h,F,G,H” to denote functions. If the relation f A x B is a function, we write it as:

f : A B If there is a way to compute yB from any given

xA, we usually write ‘f(x)’ in place of ‘y’.

Therefore, the definitions instead of…

1. xA, yB, x R y

2. xA, y1,y2B, x R y1 x R y2 y1=y2.

1. xA, yB, y = f(x)

2. xA, y1,y2B, y1=f(x) y2=f(x) y1=y2.

… can be also expressed as…

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5. The identity function.

The identity function on any given set, is a function that maps every element to itself.

idA : A A, xA, idA(x) = x

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Overview









1. g f is a function.

2. f idA idB f f3. (h g) f = h (g f)4. (g f)-1 = f-1 g-1



• Preservation under ‘’.• On inverses


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6. Definition: domain, codomain, range

Given a function: f : A B,– The set A is known as the domain of f.– The set B is known as the codomain of f.– The set { yB | xA, y=f(x)} is known as the

range of f.

A Bf

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range of f.

A Bf

Domain(f)

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range of f.

A Bf

Codomain(f)

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range of f.

A Bf

Range(f)

NOTE:

range(f) codomain(f)

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range of f.

Example 1: f : Z Z+, f(x) = |2x|

a. domain(f) = Z

b. codomain(f) = Z+

c. range(f) = positive even numbers: {2x | x Z+}

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range of f.

Example 2:

Let f : Z Z, f(x) = 2x + 1; g : Z Z, g(x) = 2x - 1

Then range(f) = range(g).

Range(f) = 2x+1 = 2(x+1) – 1 = Range(g)

xZ, f(x) = g(x+1)

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7. Definition: Image, Inverse Image.

Given a function: f : A B, and S A– The image of S is defined as:

Imgf(S) = { yB | xS, y=f(x)}

A Bf

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Imgf(S) = { yB | xS, y=f(x)}

S

Image of S

NOTE: Imgf(S) range(f) codomain(f)

A Bf

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Imgf(S) = { yB | xS, y=f(x)} Given a function: f : A B, and T B

– The inverse image of T is defined as:

InvImgf(T) = { xA | yT, y=f(x)}

Inverse image of T

TA B

f

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Imgf(S) = { yB | xS, y=f(x)} Given a function: f : A B, and T B

– The inverse image of T is defined as:

InvImgf(T) = { xA | yT, y=f(x)}

Example: f : Z Z+, f(x) = |x|

a. Imgf({10,-20}) = {10,20}

b. InvImgf({10,20}) = {10, 20, -10, -20}

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Imgf(S) = { yB | xS, y=f(x)}or

yImgf(S) iff xS and y=f(x)

Given a function: f : A B, and T B– The inverse image of T is defined as:

InvImgf(T) = { xA | yT, y=f(x)}or

xInvImgf(T) iff yT and y=f(x)

Axiomatic definition

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7.1 Theorem: Image, Inverse Image.

Given any function: f : A B, S A, T B1. S InvImgf(Imgf(S))

2. Imgf(InvImgf(T)) T

A Bf

Proof of (1): (Use diagrams to help visualise!)

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A Bf

S

Proof of (1): (Use diagrams to help visualize!)



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A Bf

S

Proof of (1):



Assume xS

Therefore xInvImgf(Imgf(S))

Then yB such that y = f(x) (Since f is a function)

Therefore yImgf(S) (Defn: yImgf(S) iff xS and y=f(x))

xy

Therefore yImgf(S) and y = f(x)

Imgf(S)

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A Bf

Imgf(S)

Proof of (1):



Assume xS

Therefore xInvImgf(Imgf(S))

Then yB such that y = f(x) (Since f is a function)

Therefore yImgf(S) (Defn: yImgf(S) iff xS and y=f(x))

Therefore yImgf(S) and y = f(x)

(Defn:xInvImgf(T) iff

yT and y=f(x))

xy

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…Proof of (2) left as an exercise…

(Again, the skill that you must pick up is that you must use diagrams wherever

possible, to help visualize the problem and use it to help you reason about proofs)



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Overview









1. g f is a function.

2. f idA idB f f3. (h g) f = h (g f)4. (g f)-1 = f-1 g-1



• Preservation under ‘’.• On inverses


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8. Operations on Functions.

A function is a (special) relation. A relation is a set (of ordered pairs). Therefore, all definitions and operations

on sets and relations are extended over to functions.

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8.1 Equality of Functions

Given 2 functions, f : A B, g : A B,

f = g iff f g and g f Or, in cases where y can be computed directly

from x:

f = g iff

xA, y1=f(x) y2=g(x) y1=y2

Which is equivalent to:

f = g iff xA, f(x) = g(x)

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8.2 Union, Intersection, Difference

Given 2 functions, f : A B, g : A B, f g, f g, fg are defined as accordingly in set theory (since functions are sets).…Or, in cases where y can be computed directly from x:– (fg)(x) = y iff y = f(x) or y g(x)

– (fg)(x) = y iff y = f(x) and y g(x)

– (fg)(x) = y iff y = f(x) and y g(x)

NOTE: f g, f g, fg need not necessarily be functions. At most, one can only conclude that f g, f g, fg are relations. In order to assert that they are functions, one must prove that the 2 conditions necessary for functions are true.

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Theorem: Given 2 functions f and g from A to B,1. If f g is a function, then f = g2. If f g is a function, then f = g3. If fg is a function, then f g =

Proof of (1): (Direct proof: f = g iff xA, f(x) = g(x)

A Bf

x

A Bg

x

A Bf g

x

y1

y2

y2

y1But fg is a function!

Defn:y1= (fg)(x) and y2= (fg)(x) then y1=y2.

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Proof of (1): (Direct proof: f = g iff xA, f(x) = g(x)Let xA.

Since f is a function, then y1B, y1= f(x), (x,y1)f

Since g is a function, then y2B, y2= g(x), (x,y2)g

Therefore (x,y1) f g and (x,y2) f g.

i.e. y1f g)(x) and y2 f g)(x)

But since fg is a function, then y1=y2 .

Therefore f(x) = g(x) (By definition of function equality)

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Proof of (3): (By contradiction)

A Bf g

x y

A B f

x y

A B g

x y

A B f-g

x

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Proof of (3): (By contradiction)

Assume that f g

Then by definition, (x,y) f g

That means that (x,y) f and (x,y) g

That means that (x,y) (f – g).

But f – g is a function! xA, yB, y = (f – g)(x). So there must be a (x,y) (f – g) =>

Contradiction!

Therefore f g .

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Proof of (2) left as an exercise

Similar to Proof of (1).

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8.3 Definition: Inverse of a function

Given a function, f : A B, the inverse of a function is defined in the same way as that of a relation:

f-1 = { (y,x) | (x,y) f }

NOTE: f-1 need not necessarily be a function. It is first and foremost, a relation. In order to assert that they are functions, one must prove that the 2 conditions necessary for functions are true. So

f-1 B x A

If f-1 is a function, then we write:

f-1 : B A

… and then say that:

f(x) = y iff f-1(y) = x

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8.3 Definition: Inverse of a function

Find the inverse of the function f(x) = (1+x)/(1-x).

Ans:– Let y = (1+x)/(1-x). Express x in terms of y.

– Therefore: y – yx = 1 + x

– Therefore: y – 1 = yx + x

– Therefore: x(y+1) = y–1

– Therefore: x = (y–1)/(y+1)

– Therefore: f-1(y) = (y–1)/(y+1)

– i.e. f-1(x) = (x–1)/(x+1)

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8.4 Definition: Composition of functions

Given 2 functions, f : A B, g : B C, then the composition of function ‘g o f’ is defined (in the same way as relation composition) as:

g o f = { (x,z) | yB, (x,y)f (y,z)g}

Or,

z = (g o f)(x) iff yB, y = f(x) z = g(y)

And since f and g are functions, y if it exists, must be unique. So:

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

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8.5 Properties of function composition

Given any functions, f : A B, g : B C, h : C D,1. g o f is a function. (o preserves function property)

2. f o idA idB o f f3. (h o g) o f = h o (g o f) (From relations: o is associative)4. (g o f)-1 = f-1 o g-1 (From relations)

Since a function is a relation, there’s no need to prove (3) and (4) since the properties of (3) and (4) are proven in the lecture on relations.

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Given any functions, f : A B, g : B C, h : C D,1. g o f is a function. ( o preserves function property)


Proof of (1): Let f : A B, g : B C. Is g o f a function?

Proof of condition 1: xA, zC, z = (g o f)(x) Let xA. Then yB, y = f(x) (Since f is a function) Also, zC, z = g(y) (Since g is a function) Therefore zC, z = g(f(x)) Therefore zC, z = (g o f)(x)

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Given any functions, f : A B, g : B C, h : C D,1. g o f is a function. (o preserves function property)


Proof of (1): Let f : A B, g : B C. Is g o f a function?

Proof of condition 2:

xA, z1=(g o f)(x) z2=(g o f)(x) z1=z2.

Let z1=(g o f)(x) z2=(g o f)(x)

z1=g(f(x)) z2=g(f(x)) (Defn of composition)

z1=g(y) z2=g(y) (Since f is a function)

z1=z2. (Since g is a function)Yes, g o f Is a function.

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Given any functions, f : A B, g : B C, h : C D,1. g o f is a function. (o preserves function property)2. f o idA idB o f f3. (h o g) o f = h o (g o f) (From relations: o is associative)4. (g o f)-1 = f-1 o g-1 (From relations)

Proof of (2):

(f o idA)(x) = f (idA(x)) (since aA, idA(a) = a)

= f(x)

(idB o f )(x)= idB(f(x)) (since bB, idB(b) = b)

= f(x)

Therefore f o idA idB o f f.

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Overview









1. G o f is a function.

2. F o idA idB o f f3. (h o g) o f = h o (g o f)4. (g o f)-1 = f-1 o g-1



• Preservation under ‘o’.• On inverses


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9. Special Kinds of functions

Three types of functions:– Injective (1-1)– Surjective (onto)– Bijective (1-1 correspondence)

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9.1 Injective (1-1) Functions

Definition: Let f : A B. f is injective (or one-to-one) iff

x1,x2A, f(x1) = f(x2) x1 = x2

A Bf

x1

x2

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x1,x2A, f(x1) = f(x2) x1 = x2

A Bf

x1

x2

An element in B can only receive from at most one element in A.

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x1,x2A, f(x1) = f(x2) x1 = x2

NOTE: This is different from saying:

x1,x2A, x1 = x2 f(x1) = f(x2)

A Bf

x1x2

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x1,x2A, f(x1) = f(x2) x1 = x2

NOTE: This is different from saying:

x1,x2A, x1 = x2 f(x1) = f(x2)

A Bf

x1x2

An element in A can only send to at most one element in B.

(2nd cond of function)

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x1,x2A, f(x1) = f(x2) x1 = x2

Q: Is f an injection?

A: No.

BA

x1

x2

f

Examples:

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x1,x2A, f(x1) = f(x2) x1 = x2


A: Yes.

A

x1

x2

Bf

Examples:

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x1,x2A, f(x1) = f(x2) x1 = x2

Examples:

f : Z Z, f(x) = x2


A: No.

A

1

-1

Bf

1F(1) = F(-1) but 1 -1

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x1,x2A, f(x1) = f(x2) x1 = x2

Examples:


A: No.

f : Z Z, f(x) =

x – 1 if x >= 0

x if x<0

A-2

Bf

0-1012

1

-1F(0) = F(-1) but 0 1

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x1,x2A, f(x1) = f(x2) x1 = x2

Examples:

f : Z Z, f(x) = 3x+2


A: Yes. Proof:

Assume f(a) = f(b)3a+2 = 3b+2a=b A Bf

-1012

-1258

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9.2 Surjective (onto) Functions

Definition: Let f : A B. f is surjective (or onto) iff

yB, xA, f(x) = y

A B

f

Again, note that this is different from saying:

xA, yB, f(x) = y

(This is 1st condition of a function)

(“Every girl is loved by some boy”)

(“Every boy loves some girl”)

(Boys) (Girls)

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yB, xA, f(x) = y Or,

Codomain(f) = Range(f)

Q: Is f a surjection?

A: No.

Examples:

A

a

b

Bf

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A: Yes.

A

a

b

Bf

Examples:

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Examples:

f : Z Z, f(x) = x2


A: No.

yB, xA, f(x) y. Take y = -1.

-1?

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Examples:

f : Z Znonneg, f(x) = x2


A: No.

yB, xA, f(x) y. Take y = 3.

3?1

24

1

68





Examples:


A: No.


0?-1

01

-1

f : Z Z, f(x) =x+1 if x>=0

x otherwise

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Examples:

f : R R, f(x) = 4x-1

Take any y, where y = 4x + 1.

There exist an x, where x = (y-1)/4 (which is in R)


A: Yes.

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Examples:

f : Z Z, f(x) = 4x-10?

-1

03

-5



A: No.

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Application– Pigeon-Hole Principle (Later)– Hashing

• Static Hashing• Dynamic Hashing

– Linear Hashing– Extendible Hashing

(Database Management Systems Course)

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9.3 Bijective Functions

Definition: Let f : A B. f is bijective (or 1-1 correspondence) iff– f is injective (1-1); and– f is surjective (onto)

Not injective, Not surjective

Injective, but not surjective

Not injective, but surjective

Injective AND surjective = bijective

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9.3 Bijective Functions

Application:– Cardinality (Next lecture)

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9.4.1 Theorem (preservation under composition)

Theorem (preservation under composition): Given f : AB and g : BC,

1. If f and g are injective, then (g o f) is injective.

2. If f and g are surjective, then (g o f) is surjective.

3. If f and g are bijective, then (g o f) is bijective.

4. The converse of the above 3 statements is not true

Proof of (1): x1,x2A, (g o f)(x1) = (g o f)(x2) x1 = x2

(g o f)(x1) = (g o f)(x2)g(f(x1))= g(f(x2)) (By defn of composition)f(x1) = f(x2) (Since g is injective)x1 = x2 (Since f is injective)

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f g

g o f

Converse of (1) is not true.

g o f is injective.

But g is not injective.

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Proof of (2): (cC, aA, (g o f)(a) = c)

Let cC

Since g is surjective, then bB, g(b) = c

Since f is surjective, then aA, f(a) = b

Therefore aA, bB, f(a) = b and g(b) = c

Therefore aA, g(f(a)) = c

Therefore aA, (g o f)(a) = c

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f g

g o f

Converse of (2) is not true.

g o f is surjective

But f is not surjective.

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Proof of (3):

Since f and g are both bijective, then by definition, f and g are both injective and surjective.

By previous two theorems, g o f is injective.and also g o f is surjective

Therefore, g o f is bijective.

(The converse is not true. It follows from the previous 2 counter-examples.)

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9.4.2 Theorem (On inverses)

Theorem: Given f : A B,

If f is a bijection, then f-1 is a function from B to A.

Proof:

(a) To show 1st condition: Every element in B must map to some element in A.

Assume any y B,

Since f is surjective (due to the fact that f is bijective), then xA such that f(x) = y.

So xA such that x = f-1(y)

A B

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9.4.2 Theorem (On inverses)

Theorem: Given f : A B,

If f is a bijection, then f-1 is a function from B to A.

Proof:

(b) To show 2nd condition: An element in B must map to at most one element in A.

Assume f-1(y) = x1 and f-1(y) = x2

Then f(x1) = y and f(x2) = y

Since f is injective (due to the fact that f is bijective), then x1 = x2

A B

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9.4.3 Theorem (On inverses)Theorem: Given f : A B,

1. If f is a bijection, then f-1 is a bijection.2. If f is a bijection, then

a. f-1 o f = idA

b. f o f-1 = idB

Proof of (1):

(a) To show is f-1 injective (1-1)

Assume f-1(y1) = f-1(y2) = x

Then x = f-1(y1) and x = f-1(y2)

Therefore f(x) = y1 and f(x) = y2

Therefore y1 = y2 (Since f is a function)

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a. f-1 o f = idA

b. f o f-1 = idB

Proof of (1):

(b) To show is f-1 surjective (onto)

(aA, bB, f-1(b) = a)

Assume aA

Since f is a function, therefore bB, f(a) = b

bB, f-1(b) = a.

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a. f-1 o f = idA

b. f o f-1 = idB

Proof of (2a):

(f-1 o f)(x) = f-1(f(x))

Let x’ = f-1(f(x)).

Then f(x’) = f(x) (Since f-1(b) = a iff f(a) = b)

So x’ = x (Since f is a bijection)

Therefore x = f-1(f(x)). Meaning that: (f-1 o f)(x) = x.

So f-1 o f = idA

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a. f-1 o f = idA

b. f o f-1 = idB

Proof of (2b):

(f o f-1)(x) = f (f-1(x))

Let y’ = f (f-1(y)).

Then f-1(y) = f-1(y’) (Since f-1(b) = a iff f(a) = b)

So y = y’ (Since f-1 is a bijection)

Therefore y = f (f-1(y)). Meaning that: (f o f-1)(y) = y.

So f o f-1 = idB

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10. Multi-argument functions

In general, a function can take in more than 1 argument.

f : (A x B) C Example:

f : (Z x Z) Z , where f(x,y) = 2x + 3y f is a function:

– Condition 1: Every element in (Z x Z) is mapped to some element in Z.

– Condition 2: The mapping is a unqiue mapping: if f(x,y) = j and f(x,y) = k, then it must be true that j = k.

86


Example:

f : (Z x Z) Z , where f(x,y) = 2x + 3y

(0,0)(0,1)(1,0)

(0,-1)

123

0-1-2

87


Multi-argument functions can be expressed as functions to functions.

Example:

f : Z (Z Z) , where f x y = 2x + 3y

012

-1 01

-103

-3

01

-125

-1

88


Multi-argument functions can be expressed as functions to functions.

Example:

f : Z (Z Z) , where f x y = 2x + 3y

The result of a applying a function to an argument is another function!!!

Expressing multi-argument functions in this way is known as ‘currying’.

Taught in courses on programming language theory

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End of Lecture

1 Annoucement n Skills you need: (1) (In Thinking) You think and move by Logic Definitions...

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