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!