Foundations of Discrete Mathematics Chapter 3 By Dr. Dalia M. Gil, Ph.D

Foundations of Discrete Mathematics

Chapter 3

By Dr. Dalia M. Gil, Ph.D.

Function

A function from a set A to a set B is a binary relation f from A to B with the property that,

For every a A, there is exactly one b B such that (a, b) f.

Function

A function f can see as a subset of A x B with the property that for each a A,

there is just one pair (a, b) in f having first coordinate a.

Function

If A = {1, 2, 3}, B = {x, y}, and

f={(1,x), (2, y), (3,x)},

then f is just the rule that associates x with 1, y with 2, and x with 3.

Function: Key Points

1. Every a in A must be the first coordinate of an ordered pair in the function.

A = {1, 2, 3} and B = {x, y}, the set g={(1,x),(3,y)} is not a function from A to B.

g contains no ordered pair with first coordinate 2.

Function: Key Points

2. Each element of A must be the first coordinate of exactly one ordered pair (no repetition).

A = {1, 2, 3} and B = {x, y},the set h={(1,x),(2,x),(3,y),(2,y)} is not a function from A to B.

2 is the first coordinate of two pairs.

Function: Example 1

Suppose A is the set of surnames of people listed in the Salt Lake City telephone directory

f = {(a, b)| a is on page n}

is a function from A to the set of natural numbers?

Function: Example 1

By definition of f, each element of A is the first coordinate of a pair in f, so

Some surnames are undoubtedly listed on a number of different page.

f is not a function.

Function

f is a function from A to B,

each b B, which is uniquely determined by the element a A,

b is denoted f(a) and called the image of a.

(a, b) f if and only if b = f(a).

Function Notation

f : A → B

It means that f is a function from A to B.

A and B are sets.

Function Notation

f : a

It means that f(a) = b a and b are elements.

b

Function Notation

f = { (1, x), (2, y), (3, x)} could be described by

1

f: 2

3

x

y

z

Function Notation

f = x2

f: R →R that associates with any x R, its square x2; that is, x x2

f = { (x, x2) | x R} as a binary relation.

Functions

Let f : A → B be a function from A to B.

The domain of f (dom f) is the set A.

The target of f is the set B.

Functions

Let f : A → B be a function from A to B.

The range or image of f (rng f) is

rng f ={b B| (a, b) f for some a A}

={b B| b = f(a) for some a A}

Onto Function

A function is onto or surjective if its range is the target, rng f = B

every b B is of the form b = f (a) for some a A

Onto Function

For any b B, the equation b = f(x) has a solution a A.

One-to-One Function

A function is one-to-one (1-1) or injective if and only if different elements of A have different images.

a1 ≠ a2 → f(a1) ≠ f(a2)

If f(a1) = f(a2), then a1 = a2

Bijective Function

A function is bijection or bijective if it is both

one-to-one and onto.

One-to-One Function

If f(a1) = f(a2), then a1 = a2

a1 ≠ a2 → f(a1) ≠ f(a2)

Example : Discrete Function

A={1, 2, 3, 4}, B ={x, y, z} andf = {(1,x),(2,y), (3, z), (4, y)}

The f is a function A → B

Domain: A and target: B

rng f = {x, y, z} = B, f is onto

f(2) =f(4) = y but 2 ≠ 4, f is not one-to-one


A={1, 2, 3}, B ={x, y, z, w} andf = {(1,w),(2,y), (3, x)}

The f is a function A → B

Domain: A and target: B

rng f = {x, y, w} ≠ B, f is not onto

f(1) ≠ f(2) ≠ f(3), f is one-to-one


A={1, 2, 3}, B ={x, y, z} and f = {(1,z),(2,y), (3, y)} and

g = {(1,z), (2,y), (3,x)}

f and g are functions A → B

dom f = dom g = A and target B rng f = {z, y} ≠ B, f is not onto rng g = {z, y, x} = B, g is onto


A={1, 2, 3}, B ={x, y, z} and f = {(1,z),(2,y), (3, y)} and

g = {(1,z), (2,y), (3,x)}

f and g are functions A → B

f(2) = f(3), 2 ≠ 3, f is not one-to-one g(1) ≠ g(2) ≠ g(3), g is one-to-one


Let f: Z → Z by f(x)= 2x – 3dom f = Z and target Z

To find rng f, note that

b rng f ↔ b = 2a – 3 for some integer a

↔ b = 2a – 3 – 1 + 1

↔ b = 2(a – 2) + 1 for some integer a

if and only if b is odd.



The range of f is the set of odd integers f ≠ Z, f is not onto

b rng f ↔ b = 2a – 3 if and only if b is odd.



f is one-to-one. if f(x1) = f(x2), then

2x1 – 3 = 2x2 – 3 and

x1 = x2


Let f: N → N by f(x)= 2x – 3

f(1) = 2(1) – 3 = -1 and -1 N

Hence, no function has been defined.

Problem about a Function

Define f: Z → Z by f(x)= x2 – 5x +5Determine whether f is one-to-one and/or

onto.

Consider f(x1) = f(x2)

x12 – 5x1 + 5 = x2

2 – 5x2 + 5

x12 – x2

2= 5x1 – 5x2 + 5 – 5

(x1 – x2) (x1 + x2) = 5(x1 – x2)


Define f: Z → Z by f(x)= x2 – 5x +5

(x1 – x2) (x1 + x2) = 5(x2 – x1)

(x1 + x2) = 5

There are solutions with x1 ≠ x2.

Any x1, x2 satisfying x1 + x2 = 5



(x1 + x2) = 5

x1 = 2, x2 =3, x1 + x2 = 5

Since

f(2) = f(3) = -1, f is not one-to-one



f(x)= x2 – 5x +5, x R, is a parabola with vertex (5/2, -5/4) any integers < -1 is not in rng f 0 is not in rng f because x2 – 5x +5 = 0

has not integer solutions.

f is not onto.

The Identity Function

For any set A, the identity function on A is the function A : A → A defined by

A(a) = a for all a A.

In terms of ordered pairs A = {(a, a) | a A }

A is read “yota sub A”


The identity function on a set A is one-to-one

If (a1) = (a2), then a1 = a2 ,

[(a1) = a1 and (a2) = a2 ],

so is one-to-one


The identity function on a set A is onto

the equation a = (x) has a solution for any a.

If x = a, then (x) = (a) = a so is onto

The Absolute Value Function

The absolute value of a number x, denoted |x|, is defined by

x if x ≥ 0 |x|= -x if x < 0

The Absolute Value Function

The domain R and range [0, )

= {y R | y ≥ 0}. It is not one-to-one

For example |2| = |-2|

The Floor Function

For any real number x, the floor of x, written x, is

the greatest integer less than or equal to x, that is, the unique integer x satisfying

x – 1 < x x

The Floor Function

x – 1 < x x

2.01 = 2,

15 = 15,

1.99 = 1,

-2.01 = -3

The Ceiling Function

For any real number x, the ceiling of x, written x, is

the least integer greater than or equal to x, that is, the unique integer x satisfying

x x < x + 1

The Ceiling Function

x x < x + 1

2.01 = 3,

15 = 15,

1.99 = 2,

-2.01 = -2

The Inverse of a Function

A function f: A → B has an inverse if and only if

the set obtained by reversing the ordered pairs of f is a function B → A.


If f: A → B has an inverse, the function f -1 = {(a, b) | (a, b) f}

is called the inverse of f.


If f: A → B has an inverse, then f -1 has an inverse that is f

( f -1 )-1= f is called the inverse of f.

If A= {1, 2, 3, 4} and B = {x, y, z, t}

f ={(1, x), (2, 2), (3, z), (4, t)}

f -1 ={(x,1), (y, 2), (z, 3), (4, t)}


A function f: A → B has an inverse, B →A if and only if f is one-to-one and onto.


For any function g (x, y) g y = g(x)

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

a= f-1(b) ↔ (b, a) f-1 ↔ (a, b) f ↔ b = f(a)

a= f -1(b) if and only if f(a) = b

a= f -1(b) ↔ f(a) = b


Example 1,

For f , = f -1(-7), then f() = 7

Example 2,

For f , f(4) = 2, then 4 = f -1(2)


If: R → R is defined by f(x) = 2x – 3 is one-to-one and onto, so an inverse function exists.

if y = f -1(x), then x = f(y) = 2y – 3

Thus, y = ½(x + 3) = f -1(x)


Let A = {x R | x 0}, B = {x R | x ≥ 0},

and define f: A → B by f(x) = x2

This squaring function with domain restricted so that is one-to-one as well as onto.

Since f is one-to-one and onto, it has an inverse.


f: R → R+ by f(x) = 3x is one-to-one and onto.Find the f-1(x)

y = f -1(x) f(y) = x 3y = x y = log3 x

f-1 (x) = log3 x


Let A = {x| x ≠ ½ } define f: A → R by f(x) = 4x / (2x – 1) is one-to-one?

Suppose f(a1) = f(a2), then4a1 / (2a1 – 1) = 4a2 / (2a2 – 1)

8a1a2 – 4a1 = 8a1a2 – 4a2

– 4a1 = – 4a2

so f is one-to-one.


Let A = {x| x ≠ ½ } define f: A → R by f(x) = 4x / (2x – 1) Find rng f

y rng f ↔ y = f(x) for some x A

↔ there is an x A | y = 4x / (2x – 1) ↔ there is an x A | 2xy – y = 4x ↔ there is an x A | x(2y – 4) = y

x = y/(2y – 4)


A = {x| x ≠ ½ }

x = y/(2y – 4), x ≠ ½ , x A

y rng f y ≠ 2, so

rng f = B = {y R| y ≠ 2}

f(x) = 4x / (2x – 1)

Let A = {x| x ≠ ½ } define f: A → R by f(x) = 4x / (2x – 1) Find the inverse

Also, dom f -1 = rng f = B

rng f -1 = dom f = A x = f(y) = 4y/(2y -1) f-1(x) = y = x / (2x – 4 )

f: A → B is one-to-one and onto. It has an inverse f -1:B → A


Let A = {x R | x 0}, B = {x R | x ≥ 0},f(x) = x2 , find the f-1(x)

y = f-1(x)

f(y) = x y2 = x y = x x = f(y) , y A, so y 0 y = -x, f-1 (x) = -x


If f: A → B is one-to-one and onto, then

f -1: B → A is also one-to-one and onto.

Composition of Functions

g ° f: A → C defined by

(g ° f) (a) = g (f (a)) for all a A

If f: A → B and g: B → C are functions, then the composition of g and f is the

function g ° f: A → C


(g ° f) (a) = g(f(a)) = g(x) = u (g ° f) (b) = g(f(b))= g(y) = w (g ° f) (c) = g(f(c))= g(x) = u

If A={a, b, c}, B={x, y}, and C = {u, v, w}, and if f: A → B and g: B → C are functions

f = {(a, x), (b, y), (c, x)}, g = {(x, u), (y, w)}


(g ° f) = {(a, u), (b, w), (c, u)}

If A={a, b, c}, B={x, y}, and C = {u, v, w}, and if f: A → B and g: B → C are functions

f = {(a, x), (b, y), (c, x)},

g = {(x, u), (y, w)}


(g ° f)(x)= g(f(x)) = g(2x – 3) = (2x – 3)2 + 1

(f ° g) (x) = f(g(x))= f(x2 + 1) = 2(x2 + 1) – 3

(g ° f) ≠ (f ° g)

If f and g are the functions R → R defined by

f(x) = 2x – 3 g(x) = x2 + 1


In the definition of g ° f, it is required that

rng f B = dom g.

Equality of Functions

Functions f and g are equal if and only if they have

• the same domain,

• the same target, and

• f (a) = g (a) for every a in the common domain.

Compositions of Functions

Function f: A → B and g: B → A are inverses if and only if

g ° f = A and f ° g= B if and only if

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

a A and b B


Show that the functions f: R → (1, ) and g: (1, ) → R defined by

f(x) = 32x + 1 and g(x) = ½ log3(x – 1 ) are inverses


f: R → (1, ), f(x) = 32x + 1 g: (1, ) → R, g(x) = ½ log3(x – 1 )

For any x R, (g ° f) (x) = g (f (x)) = g(32x + 1)

= ½ (log3[(32x + 1) – 1 ])

= ½ (log3 32x) = ½ (2x)

= x


f: R → (1, ), f(x) = 32x + 1 g: (1, ) → R, g(x) = ½ log3(x – 1 ) For any x R (1, )

(f ° g)(x) = f(g(x)) = f( ½ log3(x – 1)

= 32 ½ log3(x – 1) + 1

= 3log3(x – 1) + 1

= (x – 1 ) + 1 = x


Conclusion:

f(x) = 32x + 1 and g(x) = ½ log3(x – 1 ) are inverses because of

g ° f = A and f ° g= B g(f(a)) = a and f(g(b)) =

b

Finite and Infinite Set

A finite set is a set that is either empty or in one-to-one correspondence with the set {1, 2, …, n} of the first n natural numbers, for some n N.

A set that is not finite is called infinite.

Cardinality

The cardinality of the empty set to be 0 and write ||= 0.

If A is nonempty finite set and in one-to-one correspondence with {1, 2, …, n}, we define the cardinality of A to be n and write |A| = n.

Cardinality

If A is nonempty finite set, its elements can be labeled a1, a2, …, an for some n and the cardinality of A is n, the number of elements in A.

Cardinality

|{a, b, x}| = 3

|{a, b}| = 2

The letters of the English alphabet comprise a set of cardinality 26

|{x R| x2 + 1 = 0} | = 0

|| = 0

Cardinality

Sets A and B have the same cardinality and we write |A| = |B|,

If and only if there is a one-to-one correspondence between them,

there exists a one-to-one onto function from A to B (or from B to A).

Cardinality

|{a, b}| = |{x, y}|

Z and 2Z have the same cardinality

Any two intervals of real numbers –open, closed, finite, infinite- have the same cardinality.

Cardinality

R and R+ have the same cardinality

For f: R → R+ defined by f(x) = 2x establishes a one-to-one correspondence.

Countably

A set A is countably infinite if and only if |A| = |N|

A set A is countable if and only if it is either finite or countably infinite.

A set that is not countable is uncountable.

Countably

The symbol 0 (“aleph naught”) has been used to denote the cardinality of the natural numbers.

A countable infinite set has cardinality 0.

Countably

A subset of a countable set is countable.

The concept of same cardinality is an equivalence relation on sets, in particular, it is transitive.

Countably

Prove that the notion of same cardinality is an equivalence relation on the family of all sets.

Countably

Reflexivity: For any set A, A is one-to-one function A A,

so A has the same cardinality as itself.

Countably

Symmetry: If A and B have the same cardinality, then there is a one-to-one onto function f: A B.

a function has an inverse f-1: B A which is one-to-one and onto,

so B and A have the same cardinality.

Countably

Transitivity: Suppose A, B and C are sets such that A and B have the same cardinality and B and C have the same cardinality.

There is a one-to-one function and onto function f: A B and a one to-one onto function g: B C.

Topics covered

Functions.

Basic Terminology.

Inverses and Composition.

One-to-one correspondence and the Cardinality of a Set.

Reference

“Discrete Mathematics with Graph Theory”, Third Edition, E. Goodaire and Michael Parmenter, Pearson Prentice Hall, 2006. pp 72-97.

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Foundations of Discrete Mathematics Chapter 3 By Dr. Dalia M. Gil, Ph.D