CS 104: Discrete Mathematics

Chapter 3 :

Fundamental Structures

CS 104: Discrete Mathematics

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2

Functions (In Book: Chapter 2-sec

2.3)

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Introduction

Amal

Rawan

Maha

Sara

Safia

A

B

C

D

FS G

• In many instances we assign to each element of a set a

particular element of a second set.

• E.g. if we have a set of students S={Amal, Rawan, Maha,

Sara, Safia} and a set of grads G = { A, B, C, D, F}

we can assign a grade for each student, this assignment is

an example

of a function.

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Functions4

Definition:

• Let A and B be nonempty sets. A function f from A to B ( A

B ) is an assignment of exactly one element of B to each

element of A.

• We write f(a) = b if b is the unique element of B assigned by

the function f to the element a of A.

• Function are sometimes also called mapping or

transformations.

Functions are specified in different ways :

Explicitly state the assignment ( as example in the previous

slide)

Give a formula, e.g.

Use a computer program to specify a function

Define a function in terms of a relation from A to B

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Functions ( Cont.) 5

a

A

b=f(a)

B

f(a)

f

Definition:

• If the function f : A B, then A is the domain and B is the co-

domain of f.

• If f(a) = b, we say that b is the image of a and a is a pre-

image of b.

• The range or image of f is the set of all images of elements of

A.

• If f is a function from A to B, we say that f maps A to B.

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• To define a function, we specify its domain, codomain, and

the mapping of elements of the domain to elements in the

codomain.

• Two functions are equal when they have the same domain,

have the same codomain, and map each element in the

domain to the same element in the codomain .

• If we change a domain or codomain or a mapping of a

function, we obtain a different function.

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Example:

What are the domain, co-domain and range of the function that

assigns grades to students?

Let G be the function that assigns grades to a student in our

class.

The domain is the set {Amal, Rawan, Maha, Sara, Safia}

The co-domain is the set {A, B, C, D, F}

The range of G is the set {A, B, C, F}

Amal

Rawan

Maha

Sara

Safia

A

B

C

D

FS G

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Functions ( Cont.)8

Example :

Let R be the relation with ordered pairs (Amal,22), (Bdoor, 24),

(Sara, 21), (Dalal, 22), (Eman, 24) and (Fadwa, 22). Here each

pair consists of a graduate student and this student’s age. Specify

a function determined by this relation.

If f is a function specified by R, then : f(Amal)=22, f(Bdoor)= 24,

f(Sara)=21, f(Dala)=22, f(Eman)= 24, f(Fadwa)=22

Here f(X) is the age of x, where x is a student.

Domain {Amal, Bdoor, Sara, Dalal, Eman, Fadwa}

Co-domain { a | a>10 and a<90}

Range {21,22,24}

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Functions ( Cont.)9

Example:

Let f be the function that assigns the last two bits of a bit string of

length 2 or greater to that string. For example, f(11010)=10.

The domain of f is the set of all bit strings of length 2 or greater

both the co-domain and the range are the set {00, 01, 10, 11}

Example:

Let f: ZZ assign the square of an integer to this integer.

Domain and Co-domain of f is the set of all integers

Range of f is the set of all integers that are perfect squares {0, 1,

4, 9, …}

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Functions ( Cont.)10

The domain and codomain of functions are often specified in

computer programs

e.g.

the domain of the function is real numbers ( float ) , and the

codomain of

the function is integers.

int MyFunction ( float x )

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Definition:

Let f1 and f2 be two functions from A to R. Then f1+ f2 and f1 f2

are also functions from A to R defined by

(f1+f2) (x) = f1(x) + f2(x)

(f1 f2)(x) = f1 (x) f2(x)

Example :

Let f1 and f2 be two functions from R to R such that f1 (x) = x2

and f2 (x) = x – x2. What are the functions f1+f2 , f1 f2?

(f1+f2)(x) = f1 (x) + f2 (x) = x2 + (x – x2) = x

(f1f2)(x) = f1 (x) f2(x) = x2(x – x2) = x3 – x4

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A function f from the set A to the set B is said to be one-to-one or an

injunction, if and only if f(a) = f(b) implies that a = b for all a and b in the

domain of f. (No element of B is the image of more than one element in A)

a b ( f (a) = f (b) a=b )

Taking the contrapositive of the implication in the definition:

a b (a ≠ b f (a) ≠ f (b))

Every b B has at most one preimage.

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Example :

Determine whether the function f: {a, b, c, d} {1, 2, 3, 4, 5} with f(a)

= 4, f(b) = 5, f(c) = 1, and f(d) = 3 is injective.

The function f is one-to-one since f takes on different values at the four

elements of its domain.

Example :

Determine whether the function f : Z Z , f (x) = x2 is injective.

The function f is not one-to-one since, for instance, f(1) = f(-1) = 1, but 1

≠ -1.

Example :

Determine whether the function f (x) = x+1 from the set of

real numbers to itself is one-to-one.

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A function f from the set A to the set B is said to be onto or

surjection , if and only if for every element b B, there is an

element a A with f (a) = b. (All elements in B are

used.)

Every b B has at least one preimage.


Example :

Determine whether the function f: {a, b, c, d} {1, 2, 3} with f(a) = 3,

f(b) = 2, f(c)= 1, and f(d) = 3 is surjective.

The function f is onto, since three elements of the co-domain are images of

elements in the domain.

Example:

Determine whether the function f: Z Z, f(x) = x2 is surjective.

The function f is not onto since, e.g. there is no integer x with x2 = -1.

Example :

Determine whether the function f (x) = x+1 from the set of

integers to itself is onto.

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Functions can be both one-to-one and onto. Such functions

are called bijective.

Bijections are functions that are both injective and surjective.

Every b B has exactly one preimage.

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Example:

Determine whether the function f: {a, b, c, d} {1, 2, 3, 4} with f(a) =

4, f(b) = 2, f(c) = 1, and f(d) = 3 is bijective.

The function f is one-to-one since no two values in the domain are

assigned the same function value. It is also onto because all four

elements of the co domain are images of elements in the domain.

Hence, f is a bijection.

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Some examples18

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Relations

(In Book: Chapter 9- sec 9.1 , sec

9.2)

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Relations

The most direct way to express a relationship between

elements of two sets is to use ordered pairs made up of two

related elements.

A set of ordered pairs are called binary relations.

Definition:

Let A and B be sets. A binary relation from A to B is a subset

of A x B.

A binary relation from A to B is a set R of ordered pairs where

the first element of each ordered pair comes from A and the

second element comes from B.

a R b (a,b) R

a R b (a,b) ∉ R

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Relations ( Cont.)

Example:

Let A = Set of students; B = Set of courses, let R be the

relation that consists of pairs :

R = {(a,b) | student a is enrolled in course b}

Note that when a student is not enrolled in any course, there

will be no pairs in R that have this student as the first element.

Example :

Let A = Set of cities; B = Set of countries. Define the relation

R by specifying that (a, b) belongs to R if city a is the capital

of b.

For instance, (Riyadh, Saudi Arabia), (Delhi, India),

(Washington, USA) are in R.

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Relations ( Cont.)

Example:

Let A={0, 1, 2} and B={a, b}. {(0, a), (0, b), (1, a), (2, b)} is a

relation from A to B. This means, 0Ra, but 1Rb.

Relations can be represented in two ways - as shown in the

figure:

Graphically using arrows to represent ordered pairs.

Using a table 0.

.

a

1.

.b

2.

a b RX X

X X

012

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Relations ( Cont.)

A function f from a set A to B assigns exactly one element of B

to each element of A.

Every element of A is the first element of exactly one ordered

pair in the relation.

A relation can be used to express a one to many relationships

between the elements of the sets A and B, i.e. an element of A

may be related to more than one element of B.

A function represents a relation where exactly one element of

B is related to each element of A, i.e. every element has only

one image.

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Relations ( Cont.)

Definition:

A relation on the set A is a relation from A to A. That is, a

relation on a set A is a subset of A x A.

Example:

Let A = {1, 2, 3, 4}. Which ordered pairs are in the relation

R={(a, b) | a divides b}?

(a, b) є R if and only if a and b are positive integers not

exceeding 4 such that a divides b, we see that

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

The pairs in R are displayed graphically and in tabular form:

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Relations ( Cont.)

The pairs in R are displayed graphically and in tabular form:

1 .1

2. .2

3. .3

4. .4

1 2 3 4 RX X X X

X X X X

1234

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Relations ( Cont.)

Example :

Consider the relations on the set of integers:

R1= {(a, b) | a ≤ b}, R2={(a, b) | a > b},

R3={(a, b) | a = b or a = -b}, R4={(a, b) | a = b},

R5={(a, b) | a = b+1}, R6={(a, b) | a+b ≤ 3},

Which of these relations contain each of the pairs (1, 1), (1, 2),

(2, 1), (1, -1) and (2, 2)?

(1, 1) is in R1, R3, R4 and R6; (1, 2) is in R1and R6;

(2, 1) is in R2, R5 and R6; (1, -1) is in R2, R3 and

R6;

(2, 2) is in R1, R3 and R4.

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Relations ( Properties of Relations )

1. Reflexivity: A relation R on a set A is called reflexive if (a, a)

R for all a A.

2. Symmetry: A relation R on a set A is called symmetric if (b, a)

R whenever (a, b) R, for all a, b A.

3. Antisymmetry: A relation R on a set A is called antisymmetric

if for all a, b A, if (a, b) R and (b, a) R, then a = b.

The contrapositive is :

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4. Transitivity: A relation R on a set A is called transitive if (a, b)

R and (b, c) R imply (a, c) R, for all a, b, c A.

Remark:

The terms symmetric and antisymmetric are not opposites.

A relation can have both of these properties or may lack both of

them.

A relation can not be both symmetric and antisymmetric if it

contains some pair of the form (a,b) where a b

In antisymmetric, the only way to have a related to b and b

related to a is for a and b to be the same element.

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Example :

Which of the following relations are reflexive and symmetric?

Consider the relations on the set of integers:

R1= {(a, b) | a ≤ b}, R2={(a, b) | a > b},

R3={(a, b) | a = b or a = -b}, R4={(a, b) | a = b},

R5={(a, b) | a = b+1}, R6={(a, b) | a+b ≤

3},

The reflexive relations are R1 (because a ≤ a, for all integer a),

R3 and R4.

For each of the other relations ,it is easy to find a pair of the

form (a, a) that is not in the relation.

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The symmetric relations are R3, R4 and R6.

R3 is symmetric, for if a=b or a=-b, then b=a or b=-a.

R4 is symmetric, since a=b implies b=a.

R6 is symmetric, since a+b ≤ 3 implies b+a ≤ 3.

None of the other relations is symmetric.


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Example :

Which of the following relations are antisymmetric? Relations of

integers

R1= {(a, b) | a ≤ b}, R2={(a, b) | a > b},

R3={(a, b) | a = b or a = -b}, R4={(a, b) | a = b},

R5={(a, b) | a = b+1}, R6={(a, b) | a+b ≤ 3},

R1 is antisymmetric, since the inequalities a ≤ b and b ≤ a imply

that a = b.

R2 is antisymmetric, since it is impossible for a>b and b>a.

R4 is antisymmetric because two elements are related with

respect to R4 if and only if they are equal.

R5 is also antisymmetric, since it is impossible that a = b+1 and b

= a+1.

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Example:

Which of the following relations are transitive? Relations of

integers

R1= {(a, b) | a ≤ b}, R2={(a, b) | a > b},

R3={(a, b) | a = b or a = -b}, R4={(a, b) | a = b},

R5={(a, b) | a = b+1}, R6={(a, b) | a+b ≤ 3},

The transitive relations from Example 5 are R1, R2, R3 and R4.

R1 is transitive, since a ≤ b and b ≤ c imply a ≤ c.

R2 is transitive, since a > b and b > c imply a > c.

R3 is transitive, since a = ±b and b = ±c imply a = ±c.

R4 is transitive, since a = b and b = c imply a = c.

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Relations ( Properties of Relations )33

Example :

Consider following relations on {1, 2, 3}:

R1= {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3)},

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

R3={(2, 1), (2, 3), (3, 1)}

R4={(2, 3)},

Which of the relations are reflexive, symmetric,

antisymmetric and transitive?

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Definition:

A relation on a set A is called an equivalence

relation if it is reflexive, symmetric, and transitive.

1. Reflexive ( aA, aRa)

2. Symmetric (aRb => bRa)

3. Transitive (aRb and bRc => aRc)

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Example :

Let R be the relation on the set of real numbers such that aRb if

and only if a-b is an integer. Is R an equivalence relation?

As a-a = 0 is an integer for all real numbers a. So, aRa for all real

numbers a. Hence R is reflexive.

Let aRb, then a-b is an integer, so b-a also an integer. Hence bRa,

i.e., R is symmetric.

If aRb and bRc, then a-b and b-c are integers. So, a-c = (a-b) +

(b-c) is also an integer. Hence, aRc. Thus R is transitive.

Consequently, R is an equivalence relation

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End of Chapter 3

CS 104: Discrete Mathematics

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