Download presentation
Presentation is loading. Please wait.
1
Discrete Structures Chapter 5 Relations and Functions
Nurul Amelina Nasharuddin Multimedia Department
2
Objectives 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
3
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
4
Identity Function (pg 394)
The function 1A: A → A, defined by 1A (a) = a for all a A, is called the identity function for A It is a function that always returns the same value that was used as its argument. In terms of equations, the function is given by f(x) = x 1X(x) = x for all x in X
5
Function Equality When are f and g equal?
If f, g: A → B, we say that f and g are equal and write f = g, if f (a) = g (a) for all a A Eg: Define f: R → R and g: R → R by the following formulas: f(x) =|x| for all x R g(x) = √x2 for all x R Does f = g? Yes, |x| = √x2 for all x R
6
Composite Function Function composition is an operation for combining two functions If f: A → B and g: B → C, we define the composite function, which is denoted g o f: A → C, by (g o f )(a) = g (f(a)), for each a A
7
Example (1) Let A = {1, 2, 3, 4}, B = {a, b, c}, and C = {w, x, y, z} with f: A → B and g: B → C given by f = {(1, a), (2, a), (3, b), (4, c)} and g = {(a, x), (b, y), (c, z)}. For each element of A we find: (g o f) (1) = g (f (1))= g (a) = x (g o f) (2) = g (f (2))= g (a) = x (g o f) (3) = g (f (3))= g (b) = y (g o f) (4) = g (f (4))= g (c) = z So g o f = {(1, x), (2, x), (3, y), (4, z)}
8
Example (2) Are f o g and g o f equal?
Let f: R → R, g: R → R be defined by f (x) = x2, g(x) = x + 5. Then (g o f)(x) = g (f(x)) = g(x2) = x2 + 5, whereas (f o g)(x) = f (g(x)) = f(x+ 5) = (x+ 5)2 = x2 + 10x + 25 Here g o f: R → R and f o g: R → R, but (g o f)(1) = 6 ≠ 36 = (f o g)(1), so even though both composites f o g and g o f can be formed, we do not have f o g = g o f
9
Example (3) Let f, g, h: R R, where f(x) = x2, g(x) = x +5 and h(x) = Then, ((h o g) o f)(x) = (h o g)(f(x)) = (h o g)(x2) = h(g(x2)) = h(x2 + 5) =
10
Example (4) Let f, g, h: R R, where f(x) = x2, g(x) = x +5 and h(x) = Find (h o (g o f))(x). Is (h o (g o f)) = ((h o g) o f)? Yes. Using the definition of the composite function, we find that ((h o g) o f) (x) = (h o g) (f (x)) = h ( g (f (x))), whereas (h o (g o f )) (x) = h ((g o f ) (x)) = h (g (f (x))).
11
Associative Property Since ((h o g) o f)(x) = h (g(f(x))) = (h o (g o f))(x), for each x in A, it now follows that (h o g) o f = h o (g o f)
12
Composition with an Identity Function
If f is a function from a set A to a set B, and 1A is the identity function on A and 1B is the identity function on B, then f o 1A = f = 1B o f
13
Example (1) Let X = {a,b,c,d} and Y = {u,v,w} and f = {(a, u), (b, v), (c, v), (d, u)}. Find f o 1X and 1Y o f (f o 1x)(a) = f(1x(a)) =f(a) = u (f o 1x)(b) = f(1x(b)) =f(b) = v (f o 1x)(c) = f(1x(c)) =f(c) = v (f o 1x)(d) = f(1x(d)) =f(d) = u So, (f o 1x)(x) = f(x) Find (1Y o f ) and show that for (1Y o f)(x) = f(x)
14
Properties of Composite Function
Let f: A → B and g: B → C a) If f and g are one-to-one, then g o f is one-to-one b) If f and g are onto, then g o f is onto
15
Converse of a Function Since a function is also a relation, first let us see the converse of a relation For sets A, B, if R is a relation from A to B, then the converse of R, denoted Rc, is the relation from B to A defined by R = {(b, a)|(a, b) R Eg: A = {1, 2, 3, 4}, B = {w, x, y}, and R = {(1, w), (2, w), (3, x)} then Rc = {(w, 1), (w, 2), (x, 3)}, a relation from B to A
16
Example (1) For A = {1, 2, 3} and B = {w, x, y}, let f: A → B be given by f = {(1, w), (2, x), (3, y)} Then f c = {(w, 1), (x, 2), (y, 3)} is a function from B to A, and we observe that f c o f = 1A and f o f c = 1B f c o f(1) = f c(w) = 1 f c o f(2) = f c(x) = 2 f c o f(3) = f c(y) = 3 f c o f = {(1,1), (2,2), (3,3)} = 1A
17
Invertibility of a Function
If f: A → B, then f is said to be invertible if there is a function g: B → A such that g o f = 1A and f o g = 1B A function f: A → B is invertible iff it is one-to-one and onto
18
Example (1) Let f, g: R → R be defined by f(x) = 2x + 5,
g(x) = (1/2)(x – 5). Then (g o f)(x) = g(f(x)) = g(2x + 5) = (1/2)[(2x + 5) – 5] = x, and (f o g )(x) = f(g (x)) = f((1/2)(x –5)) =2[(1/2)(x– 5)] + 5 = x So f o g = 1R and g o f = 1R. Consequently, f and g are both invertible functions
19
f-1(y) = that unique element x in X such that f(x) equals y
Inverse Functions Suppose f: X Y is a one-to-one correspondence. Then inverse function, f-1: Y X that is defined as follows: Given any element in Y, f-1(y) = that unique element x in X such that f(x) equals y In other words, f-1(y) = x y = f(x) x y=f(x) f f -1
20
Finding an Inverse Function
Find the inverse function for f(x) = 4x – 1for all real numbers x By definition of f-1(y) = that unique real number x such that f(x) = y f(x) = y 4x - 1= y x = (y + 1)/4 f -1(y) = x, hence f -1(y) = (y + 1)/4 Rename y by x, f -1(x) = (x + 1)/4
22
Quiz 4A Let f: Z→Z be the successor function and let g: Z→Z be the squaring function. Then f(n) = n + 1 for all n Z and g(n) = n2 for all n Z. (a) Find the compositions g o f and f o g. (b) Is g o f = f o g? Explain. Let X = {a,b,c,d} and Y= {u,v,w}, and suppose f: X → Y is given by {(a, u), (b, v), (c, v), (d, u)}. Find f o 1X and 1Y o f. f: R → R is defined by f(x) = 3x + 5. Find its inverse function. Send your answers in the next class!
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.