Mathematical Foundation of Computer Science

DHARM

16 MATHEMATICAL FOUNDATION OF COMPUTER SCIENCE

Thus,

1. | f : X → Y | = 27, when f is arbitrary.


2. For surjective mapping img (f) = {a, b, c}. So, we have the last column and ‘abc’
   gives the | f : X → Y | = 6.


3. Since, the last column together with ‘abc’ gives that each element of X has a unique
   image in Y hence, | f : X → Y | = 6.


4. The last column together with ‘abc’ gives the bijective mapping, so | f : X → Y | = 6.


5. The monotone mapping are those in the first two columns, so | f : X → Y | = 10.


6. The antitone mapping are found in the columns three, four, fifth, and six are
   | f : X → Y | = 2 + 3 + 0 + 1 = 6.

1.5.2 Composition of Functions....................................................................................

Given two function f : X → Y and g : Y → Z, then g ◊ f is called composite function, where,
g ◊ f = g(f(x)) = {(x, z)/x ∈ X and z ∈ Z and (y = f(x) and z = g(y)) for any y ∈ Y}.
Therefore, the necessary condition for g ◊ f is, img (f) ⊆ arg (g), Otherwise g ◊ f = ∅.
Conversely, f ◊ g may or may not exist. It exists if and only if img (g) ⊆ arg (f).

Example 1.7. Let f : R → R; f(x) = – x^2 and g : R+ → R+; g(x) = x where R is the set of real
numbers and R+ is the set of positive real numbers. Determine f ◊ g and g ◊ f.

Sol. We have mapping f =

...... , , , , , ...

...... , , , , , ...

−−

−−

F

H

I

K

21012

41014 and g =

012

012

, , , ......

, , ,......

F

HG

I

KJ

To determine f ◊ g we must have img (g) ⊆ arg (f), because R+ ⊆ R.

Since, f ◊ g = f(g(x)) = f( x) = – ( x)^2 = – x.

Therefore, f ◊ g = {(x, – x)/x ∈ R};
Similarly to determine g ◊ f; since img (f) ⊆/ arg (f) because img (f) ∈ R and arg (g) ∈ R+ so R
⊆/ R+. Hence, g ◊ f does not exist.
Example 1.8. Let a function f : R → R, where f(x) = x^3 – 2 and R is the set of real numbers , find
f–1. Also show that f ◊ f–1 = I.
Sol. Given mapping is represented by the expression, i.e.

f =

... ...

... ...

x

x^3 − (^2) x
F
H
I
K ∈R ⇒ f =
... , , , , ...
... , , , , ...
−−
− −−−
F
H
I
K
21012
10 3 2 1 6
Since f : R → R is surjective, so f –1 exists, i.e.
f –1: R → R
Since, f(x) = x^3 – 2 = y (assume); then f –1 (y) = x.
⇒ x = (y + 2)1/3
⇒ y = (x + 2)1/3 [replace x and y]
Therefore, inverse of f(x) is (x + 2)1/3.
Let (x + 2)1/3 = g(x),
So, f ◊ f–1 = f ◊ g = f(g(x))
= f((x + 2)1/3)
= ((x + 2)1/3)^3 – 2;