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- On completion of this chapter, student should be able to:
- Define a relation and function
- Determine the type of function (one-to-one, onto, one-to-one correspondence)
- Find a composite function
- Find an inverse function

Outline

- Cartesian products and relations
- Functions: Plain, one-to-one, onto
- Function composition and inverse functions
- Functions for computer science
- Properties of relations
- Computer recognition: Zero-one matrices and directed graphs
- Use in database example

- For sets A, B, the Cartesian product, or cross product, of A and B is denoted by A × B and equals {(a, b) | a A, b B}
- Elements of A × B are ordered pairs. For (a, b), (c, d) A × B , (a, b) = (c, d) if and only if a = c and b = d

Properties:

If A, B are finite, it follows from the rule of product that |A × B| = |A||B|

Although we generally will not have A × B = B × A, we will have |A×B|=|B×A|

Let A = {2, 3, 4}, B = {4, 5}. Then

a) A × B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}

b) B × A = {(4, 2), (4, 3), (4, 4), (5, 2), (5, 3), (5, 4)}

c) B2 = B × B = {(4, 4), (4, 5), (5, 4), (5, 5)}

d) B3 = B × B × B = {(a, b, c) | a, b, c B}; for instance, (4, 5, 5) B3

An experiment E is conducted as follows:

A single dice is rolled and its outcome noted,

and then a coin is flipped and its outcome noted.

Determine a sample space S for E

S1={1, 2, 3, 4, 5, 6} be a sample space dice.

S2= {H, T} be a sample space coin.

Then S = S1 × S2 is a sample space for E.

At the Wimbledon Tennis Championships, women play at most three sets in a match

The winner is the first to win two sets. If we let N and E denote the two players, the tree diagram indicates the six ways in which this match can be won

For example, the starred line segment (edge) indicates that player E won the first set

The double starred edge indicates that player N has won the match by winning the first and third sets

- Let A = {0,1,2}, B = {1,2,3}. A x B = {(0,1), (0,2), (0,3), (1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
- Let say an element x in A is related to an element y in B iff x is less than y. x R y: x is related to y
- 0 R 1, 0 R 2, 0 R 3, 1 R 2, 1 R 3, 2 R 3
- The set of all ordered pair in A x B where elements are related {(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)}

- For sets A, B, a (binary) relation R from A to B is a subset of A × B. Any subset of A × A is called a (binary) relation on A
- Given an ordered pair (a, b) in A x B, x is related to y by R (x R y) iff (x, y) is in R
- In general, for finite sets A, B with |A| = m and |B|= n, there are 2mn relations from A to B, including the empty relation as well as the relation A × B itself

Let A = {2, 3, 4}, B = {4, 5}. Then

A × B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}.

The following are some of the relations from A to B.

{(2, 4)}

{(2, 4), (2, 5)}

{(2, 4), (3, 4), (4, 4)}

{(2, 4), (3, 4), (4, 5)}

A × B

Since |A × B| = 6, there are 26 possible relations from A to Β (for there are 26 possible subsets of A × B )

Let A = {1,2}, B = {1,2,3} and define a binary relation from A to be as follows:

Given any (x,y) A x B, (x,y) R x – y is even

State explicitly which ordered pairs are in A x B and which are in R

Is 1 R 3?

Is 2 R 3?

Is 2 R 2?

A x B= {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)} and R, when x – y is even = {(1,1), (1,3), (2,2)} A x B

Is 1 R 3? Yes

Is 2 R 3? No

Is 2 R 2? Yes

(1,1) R because 1 – 1= 0 is even

(1,2) R because 2 – 1 = 1 is not even

Let B={1,2} and A=P(B) = {,{1},{2},{1,2}}

|A×A| = 4.4 = 16

A×A = {(∅,∅),(∅,{1}),(∅,{2}),(∅,{1,2}),

({1},∅), ({1},{1}), ({1},{2}), ({1},{1,2})

({2},∅),({2},{1}), ({2},{2}),({2},{1,2})

({1,2},∅),({1,2},{1}),({1,2},{2}, ({1,2},{1,2})}

The following is an example of a relation on A:

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

- With A = Z+ (set of positive integers), we may define a relation R on set A as {(x, y) | x ≤ y}
- This is the familiar “is less than or equal to” relation for the set of positive integers
- It can be represented graphically as the set of points, with positive integer components, located on or above the line y = x in the Euclidean plane, as partially shown in the figure below

(7, 7), (7, 11) R (8, 2) R

(7, 11) R or 7 R 11 (infix notation)

- Let A = {1,2,3}, B = {1,3,5}
- For all x A and y B, relations S and T
- (x,y) S x < y
- T = {(2,1), (2,5)}

- For nonempty sets A and B,
- A function, or mapping, f from A to B, denoted f: A B, is a relation from A to B in which every element of A appears exactly once as the first component of an ordered pair in the relation
- Sample functions:
- f : R R, f(x) = x2
- f : Z Z, f(x) = x + 1
- f : Q Z, f(x) = 2

- A function f from a non-empty set A to a set B is a relation from A to B satisfying the following two properties:
- x A, y B such that (x,y) f
- (x, y), (x, y’) f, y = y’
- The 1st property says every x A is related to at least one y B
- The 2nd property says each x A is related to at most one y B
- That is, a relation from A to B is a function from A to B if and only if every x A is related to exactly one y B

Let A = {1,2,3}, B = {7,8,9}

g = {(1,8), (2,9), (3,9), (3,10)} A x B is not a function from A to B: (3,9), (3,10) g but 9 10. Relation g fails to be a function because 3 A is related to two (distinct) elements 9, 10 B

h = {(1,9), (2,10), (3,9)} A x B is a function from A to B. Relation h is a function because each element of A is related to exactly one element in B

- We often write f(a) = b when (a, b) is an ordered pair in the function f. For (a, b) f, b is called the image of a under f, whereas a is a preimage (inverse image) of b

The arrow diagram of a function from A to B has the characteristic that there is exactly one arrow shooting out from every element of A

However, a element of B can be hit by no arrows, one arrow, or many arrows

- For the function f: A → B, A is called the domain of f and B the codomain of f
- The subset of B consisting of those elements that appear as second components in the ordered pairs of f is called the range of f and is also denoted by f (A) because it is the set of images (of the elements of A) under f
- Eg: Let A = {1, 2, 3}, B = {w, x, y, z }, f={(1, w), (2, x), (3, x)}
- Domain of f = {1,2,3}, the codomain of f = {w, x, y, z}, and the range of f = f (A) = {w, x}

Interesting Functions in Computer Science

Greatest integer function, or floor function:

This function f: R → Z, is given by f(x) = x = the greatest integer n less than or equal to x, n x n + 1

Consequently, if x is a real number and n is an integer, then f(x) = x = is the integer to the immediate left of x on the real number line. For this function, we find that

1) 3.8 = 3, 3 = 3, –3.8 = –4, –3 = –3;

2) 7.1 + 8.2 = 15.3 = 15 = 7 + 8 = 7.1 + 8.2

3) 7.7 + 8.4 = 16.1 = 16 ≠ 15 = 7 + 8 = 7.7 + 8.4

Interesting Functions in Computer Science

Ceiling function:

This function g: R → Z, is given by g(x) = x = the least integer greater than or equal to x, n x n + 1

Consequently, if x is a real number and n is an integer, then g(x) = x = is the integer to the immediate right of x on the real number line. For this function, we find that

1) 3 = 3, 3.01 = 3.7 = 4 = 4 , –3.01 = –3.7 = –3;

2) 3.6 + 4.5 = 8.1 = 9 = 4 + 5 = 3.6 + 4.5

3) 3.3 + 4.2 = 7.5 = 8 ≠ 9 = 4 + 5 = 3.3 + 4.2

Interesting Functions in Computer Science

Trunc function (for truncation):

valued function defined on R. This function deletes the fractional part of a real number

For example, trunc(3.78) = 3, trunc(5) = 5, trunc(–7.22) = –7 Note that trunc(3.78) = 3.78 = 3 while trunc(–3.78) = –3.78 = –3

For general case, let A, B be nonempty sets with |A| = m, |B| = n. Consequently, If A = {a1, a2, …, am} and B={b1,b2,…,bn}, then a typical function f: A → B can be described by

{(a1, x1), (a2, x2), (a3, x3), …, (am, xm)} – m ordered pairs.

x1 can selected from any of the n elements of B

x2 “

…………………..

xm “

In this way, using the rule of product, there are nm = |B||A| functions from A to B

30

- Let A = {1, 2, 3}, B = {w, x, y, z}, f = {(1, w), (2, x), (3, x)}
- There are 24.3=212 = 4096 relations from A to B
- We have examined one function among these relations, and now we wish to count the total number of functions from A to B
- Therefore, there are 43 = |B||A| = 64 functions from A to B, and 34 = |A||B| = 81 functions from B to A

31

- Two important properties that functions may satisfy:
- The property of being one-to-one and
- The property of being onto
- Functions that satisfy both properties are called one-to-one correspondences or one-to-one onto functions

32

- Let f be a function from A to B. f is called one-to-one, or injective, iff for all elements x1 and x2 in A
- If f(x1) = f(x2), then x1 = x2
- or, equivalently if x1 x2, then f(x1) f(x2)
- Each element of B appears atmost once as the image of an element of A

33

- If f: A → B is one-to-one, with A, B finite, we must have |A|≤|B|
- For arbitrary sets A, B, f: A → B is one-to-one if and only if for all, a1, a2 A,
- f (a1) = f (a2) a1 = a2

36

Identifying One-to-one Functions Defined on Finite Sets

- Let X = {1,2,3} and Y = {a,b,c,d}
- Define H: X Y as follows: H(1) = c, H(2) = a, H(3) = d. Is H one-to-one?
- Define K: X Y as follows: K(1) = d, K(2) = b, K(3) = d. Is K one-to-one?

37

Identifying One-to-one Functions Defined on Infinite Sets

- Suppose f is a function defined on an infinite set X. By definition, f is one-to-one iff the following is true:
- x1, x2 X, if f(x1) = f(x2), then x1 = x2
- Suppose x1 and x2 are elements of X such that
- f(x1) = f(x2)
- (2) Show that x1 = x2

38

Consider the function f: R→ R where f (x) = 3x + 7

for all x R

Then for all x1 , x2 , R, we find that

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

3x1 = 3x2 (minus both side with 7)

x1 = x2 , (dividing both side with 3)

so the given function f is one-to-one

39

On the other hand, suppose that g: R → R is the function defined by g (x) = x4 – x for each real number x

Let x1 = 0 and x2 =1.Then

g(x1) = g(0) = (0)4 – 0 = 0

g(x2) = g(1) = (1)4 – (1) = 1 – 1 = 0

Hence g(x1) = g(x2) but x1 x2 (0 ≠ 1) – that is, g is not one to-one because there exist real numbers x1, x2 where g (x1) = g (x2) but x1 x2

40

Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5}

The function f = {(1, 1), (2, 3), (3, 4)} is a one-to-one function from A to B;

g = {(1, 1), (2, 3), (3, 3)} is a function from A to B, but fails to be one-to-one because g(2) = g(3) = 3 but 2 ≠ 3

For A, B in the above example, there are 215 relations from A to B and 53 of these are functions from A to B.

The next question we want to answer is how many functions f: A → B are one-to-one

41

Calculate Total No of One-to-one Functions

- With
- A = {a1, a2, a3, …, am},
- B = {b1, b2, b3, …, bn}, and m ≤ n ,
- a one-to-one function f: A → B has the form
- {(a1, x1), (a2, x2), (a3, x3), …, (am, xm)},
- Where there are
- n choices for x1
- n – 1 choices for x2
- n – 2 choices for x3
- ………..
- n – m+1 choices for xm.,
- The number of one-to-one functions from A to B is
- n(n-1)(n-2)…(n-m+1)= n!/(n-m)! = P(n,m)= P(|B|,|A|)

42

- Consequently, for A, B where A = {1, 2, 3} and B = {1, 2, 3, 4, 5}, there are
- P(5,3) = P(|B|,|A|)
- =5 . 4 . 3
- = 60 one-to-one functions f: A → B.

43

- A function f: A→ B is called onto, or surjective, if
- f (A) = B – that is, if for all b B there is at least one a A with f (a) = b

44

Identifying Onto Functions Defined on Finite Sets

- Let X = {1,2,3,4} and Y = {a,b,c}
- Define H: X Y as follows: H(1) = c, H(2) = a, H(3) = c, H(4) = b. Is H onto?
- Define K: X Y as follows: K(1) = c, K(2) = b, K(3) = b, K(4) = c. Is K onto?

46

Identifying Onto Functions Defined on Infinite Sets

- Suppose f is a function from a set X to a set Y, and suppose Y is infinite. By definition, f is onto iff the following is true:
- y Y, x X such that f(x) = y
- Suppose that y is any element of Y
- (2) Show that there is an element of X with f(x) = y

47

The function f: R → R defined by f(x) = x3 is an onto function

If r is any real number in the codomain of f, then the real number 3√r is in the domain of f and

f(3√r) = (3√r)3 = r

E.g. f(3) = 27, f(-3) = -27

Hence the codomain of f = R = range of f, and the function f is onto

48

- The function g: R → R, where g(x) = x2 for each real number x, is not an onto function
- In this case, no negative real number appears in the range of g
- For example, for –9 to be in the range of g, we would have to be able to find a real number r with g(r) = r2 = –9
- Note, however, that the function h: R → [ 0, +∞ ) defined by h(x) = x2 is an onto function

49

- Consider the function f: Z → Z, where f(x) = 3x + 1 for each x Z
- Here the range of f = {…, –8, –5, –2, 1, 4, 7, …} Z, so f is not an onto function
- E.g. f(x) = 3x + 1 = 8 then x = 7/3
- Rational number 7/3 is not an integer –so there is no x in the domain Z with f(x) = 8

50

On the other hand, each of the functions

1) g: Q → Q, where g(x) = 3x + 1 for x Q; and

2) h: R → R, where h(x) = 3x + 1 for x R

is an onto function (Q is a set of rational numbers: a/b)

Furthermore, 3x1 + 1 = 3x2 + 1 3x1= 3x2 x1 = x2, regardless of whether x1 and x2 are integers, rational numbers, or real numbers

Consequently, all three of the functions f, g and h are one-to-one

51

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

f1 = {(1, z), (2, y), (3, x), (4, y)} and

f2 = {(1, x), (2, x), (3, y), (4, z)}

are both functions from A onto B

However, the function g = {(1, x), (2, x), (3, y), (4,y)} is not onto, because g(A) = {x, y} B (no z!)

If A, B are finite sets, then for an onto function f: A → B to possibly exist we must have |A| ≥ |B| where

|A|= m ≥ n = |B|

52

- If f: A → B, then f is said to be bijective, or to be a one-to-one correspondences, if f is both one-to-one and onto.
- Eg: If A = {1, 2, 3, 4} and B = {w, x, y, z}, then
- f = {(1,w),(2,x),(3,y),(4, z)} is a one-to-one correspondence from A (on) to B, Why?
- Ans: f is one-to-one (every element of B appear at most once), and f is onto (f(A) = B)

53

- Let A = {1, 2, 3, 4} and B = {w, x, y, z}, and g = {(w, 1), (x, 2), (y, 3), (z, 4)}. Is g a one-to-one correspondence from B (on) to A?

54

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