FUNCTIONS:
- A
function f from a set X to a set Y is a relation from X to Y such that
x Î X is related to one
and only one y Î Y
- X is called
the domain & Y is called the range. We say x is mapped into y.
- f = {(a, 3),
(b, 3), (c, 5), (d, 1)}
- Functions are described as:
- Set of ordered pairs, example f as
given before
- Using a formula such as f(x) =
expression, example : f (x) = 2x + 2
- Illustration as in the example below
FUNCTIONS AS
RELATIONS:
Only one-to-one
and many-to-one relations are considered as a function, ie, a function f from
set X to set Y as shown above assigns exactly one element of set Y to elements
of set X denoted by an ordered pair of (x,y) such that y= f(x), where x∊X and y∊Y.
SOME FUNCTION
TERMINOLOGY:
•
If f:X®Y, and f(x)=y (where xÎX & yÎY), then:
– X is the domain of f.
– Y is the codomain of f.
– y is the image of a under f.
– x is a pre-image of b under f.
•
In
general, y may have more than one pre-image.
– The range RÍB of f is
{y | $x f
(x)= y }.
Range vs
Codomain Example:
•
Suppose
that: “f is a function mapping students in this class to the set of grades
{A,B,C,D,E}.”
•
At
this point, you know f ’s codomain is: {A,B,C,D,E} , and its range is unknown!
•
Suppose
the grades turn out all As and Bs.
•
Then
the range of f is {A,B} , but its codomain is still {A,B,C,D,E}! .
Let’s Try~
List down which is
function and not function.
C)
D)
BOOLEAN FUNCTION
DEFINITON
Boolean algebra defines operations for
the set {0, 1}
Commonly used Boolean operation.
DEGREE OF BOOLEAN
B n
= {(x1, x2, ..., xn) | xi ∈{0, 1} for 1 ≤ i ≤ n} is the set of all n-tuples of 0s and
1s
Example
F(x, y)= B
2 (2 operands)
2 n = {(0, 0), (0, 1), (1, 0), (1,
1)}
2 2 n = 16 possible functions to be
derived
16 Possible
Boolean functions to be derived here…
BOOLEAN IDENTITIES
PROPERTIES OF FUNCTIONS:
1. ONE-TO-ONE FUnction
A function is
called a one-to-one or (injective) if exactly each element in the domain
of the function are related to one domain in the range.
2. ONTO function
A function is
called an onto or (surjective) if all the elements of the range is
related to at least one element in the domain.
3. ONE-TO-ONE CORRESPONDENCE
•
A
function is bijective if it is one-to-one onto. (also
known as one-to-one correspondence)
•
Each
element in the domain is only mapped to
one element in the range.
•
All
the elements in the range is related by only one element in the domain.
INVERSE FUNCTIONS AND COMPOSITION FUNCTIONS
Inverse
Function
• Let,
A function f be a one-to-one correspondence from set A to set B. The inverse
function of f, denoted as f-1 is the function that assigns to an
element b belonging to B the unique element a in A such that f(a) = b.
• Hence, the inverse can be written as f-1(b)=a
The function f and its inverse f-1
• Remark: The inverse function of f is true only when f is an
one-to-one correspondence.
EXAMPLE:
Let f be the function from
{ali, baba, chong} to {cina, malay, india} such that f (ali) = malay, f (baba)
= india, and f (chong) = cina.
Is f invertible, and if it is,
what is its inverse?
Answer:
The function f is invertible
because it is a one-to-one correspondence. The inverse
function f−1 reverses
the correspondence given by f , so
f −1 (cina) = chong,
f −1 (malay) = ali,
f −1 (india) = baba.
Composite
Function
• Let,
A function g from set A to set B and another function f from set B to set
C. The composition function between function g and f, denoted for all a∊A by f o g, is the
function which maps the elements in set A to set C through f and g.
• Can also be written as (f o g)(a) = f (g(a))
f(g(a)) is the composite function of y=g(a) and then f(y)
Remark
: The commutative law does not apply for composition
of
f(x) and g(x), that is
f(g(x))≠ g(f(x)), unless f(x) = g(x).
EXAMPLE:
Let g be the function
from the set {a, b, c} to itself such that g(a) = b, g(b)
= c, and g(c) = a.
Let f be the function
from the set {a, b, c} to the set {1, 2, 3} such that f
(a) = 3, f (b) = 2, and
f (c) = 1. What is the composition of f and g,
and what is the composition of g and f ?
Answer:
The composition f ◦ g is
defined by (f ◦ g)(a) = f (g(a)) =
f (b) = 2,
(f ◦ g) (b) =
f (g(b)) = f (c) = 1, and (f ◦ g)(c) = f
(g(c)) = f (a) = 3.
Note that g ◦ f is not
defined, because the range of f is not a subset of the domain of g.
p/s: You can get the note from this link --> Chap 2: Set Theory (Part 6)
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