Relations and functions

Q Define function.

Q What is domain and range of a function?
Ans Domain means all possible input values and range is all possible output values.

Let us take an example

y=f(x)=4x

all the possible values or given value of x is called domain
and all possible respective output values is called range

If x={1,2,3} is domain
then y=f(x)={4,8,12} is range

Q Can a function in the form of P/Q have zero in denominator?

No

Q Function with root (root cannot be negative)

Q What is relation?

Let us understand this with an example

A={3,5} B={6,10,9,15,12,20}

Now the cartesian product of the sets is

A X B= {(3,6),(3,10),(3,9),(3,15),(3,12),(3,20),(5,6),(5,10),(5,9),(5,15),(5,12),(5,15)}

Let R be a relation ------> For all(a,b) satisfies R

R-->element b of set :B is multiple of  element a in set A
R-->{a,b}:b is a multiple of a, a belongs to A, b belongs to B
therefore  relation gives us a set of ordered pair that satisfies a relationship

Now herfe (3,6),(3,9),(3,12),(5,10),(5,15),(5,20) satisfy relation R
R'={(3,6),(3,9),(3,12),(5,10),(5,15),(5,20)}

Q Find image, domain , range and co domain in above example?

Ref to last question example
Image: For every (a,b) belongs to R'  ,  b is an image of its a where a belongs to A and b belongs to B
above 6 is an image of 3, 9 is a image of 3 etc

Domain
a of (a,b) in R'
here domain is (3,5)

Range  is b of (a,b) in R'
here range is (6,9,12,10,15,20)


Co domain is whole B Set
co domain is (6,10,9,15,12,20)

From above Range is a subset of Co Domain.

Q What are the different types of relations?

Ans  The different types of relations are

a) Empty relation

b) Universal relation

c) Reflexive, symmetric and transitive

d) Equivalence relation


Empty and Universal relations are called trivial relation

Empty relation
A={0,1,2,3,4,5}
R={(a,b):a+b>50} a,b belongs to A

Here condition in relation is not satisfied and hence the relation is
an empty relation.



Universal relation

A={0,1,2,3,4}
R={(a,b):a+b>0} a,b belongs to A

since for all elements in set A the condition in the relation is satisfied
therefore R is said to have universal relation



Let us understand   through an example

Reflexive

P={p,q}
Possible reflexive pattern is
{(p,p),(q,q)}

A={1,2,3}

i){(1,1),(2,2),(3,3)} is a reflexive set
ii) {(1,1),(2,2),(3,3),(1,2),(2,3)} is also a reflexive set
iii){(1,1),(1,2),(3,3)} is not a reflexive set
iv){ } empty set is also not reflexive set

Symmetric
P={p,q}
{(p,p},(q,q),(p,q),(q,p)}
A reflexive relation is always symmetric but all symmetric relation need not be reflexive

A={1,2,3,4}

i) { } empty set is symmetric
ii){1,2} is not symmetric
iii){(1,1),(2,2),(1,2),(2,1)} is symmetric

Transitive relation

A={1,2,3}

i){1,1} is a transitive relation
ii) {(1,2),(2,3)} it is not a transitive relation
iii){(1,1),(2,3),(3,1),(2,1)} is a transitive relation


Equivalence relation:  For a set A, where A={a,b,c} relation 'R' is an equivalent  relation
in set 'A' if and only if

i) R is reflexive ie for all a belongs to R,(a,a) belongs to R
ii) R is symmetric ie (a,b) belongs then (b,a) belongs R
iii) R is transitive  ie (a,b) belongs to R and (b,c) belongs to R then
(a,c) belongs to R

One to One Function(Injective Function) Here for  each element of set A there is  only one image/distinct element in set B

Consider if a1 ∈ A and a2 ∈ B, f is defined as f: A → B such that f (a1) = f (a2)
If f(a1) =  f(a2)
 then (a1) =  (a2)

Many to one function : If two or more different elements of a A have the same image in B. It is called many to one function.

Onto Function(surjective function) If each element of B has its preimage in A. The function is called onto or surjective function

Bijective Function :One – One and Onto Function

A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function. In other words, the function f associates each element of A with a distinct element of B and every element of B has a pre-image in A.

In mathematics, the composition of a function is a stepwise application. For example, the function f: A→ B & g: B→ C can be composed to form a function which maps x in A to g(f(x)) in C. All sets are non-empty sets. A composite function is denoted by (g o f) (x) = g (f(x)). The notation g o f is read as “g of f”.

Consider the functions f: A→B and g: B→C. f = {1, 2, 3, 4, 5}→ {1, 4, 9, 16, 25} and g = {1, 4, 9, 16, 25} → {2, 8, 18, 32, 50}. A = {1, 2, 3, 4, 5}, B = {16, 4, 25, 1, 9}, C = {32, 18, 8, 50, 2}.Here, g o f = {(1, 2), (2, 8), (3, 18), (4, 32), (5, 50)}.

The composition of functions is associative in nature i.e., g o f = f o g. It is necessary that the functions are one-one and onto for a composition of functions.

Invertible Function

A function is invertible if on reversing the order of mapping we get the input as the new output. In other words, if a function, f whose domain is in set A and image in set B is invertible if f-1 has its domain in B and image in A.

f(x) = y ⇔ f-1 (y) = x.

Not all functions have an inverse. For a function to have an inverse, each element b∈B must not have more than one a ∈ A. The function must be an Injective function. Also, every element of B must be mapped with that of A. The function must be a Surjective function. It is necessary that the function is one-one and onto to be invertible, and vice-versa.

It is interesting to know the composition of a function and its inverse returns the element of the domain.

f-1 o f = f -1 (f(x)) = x                                                                 

Problem: If f: A → B, f(x) = y = x2 and g: B→C, g(y) = z = y + 2 find g o f.
Given A = {1, 2, 3, 4, 5}, B = {1, 4, 9, 16, 25}, C = {2, 6, 11, 18, 27}.

Solution: g o f(x) = g(f(x))
g(f(1)) = g(1) = 2, g(f(2)) = g(4) = 6, g(f(3)) = g(9) = 11, g(f(4)) = g(16) = 18, g(f(5)) = g(25) = 27.

Problem: Write the inverse of the above g o f.

Solution: (g o f) -1 = f-1(g-1(z))
f-1(g-1(z)) = f-1(g-1(2)) = f-1(1) = 1, f-1(g-1(6)) = f-1(4) = 2, f-1(g-1(11)) = f-1(9) = 3, f-1(g-1(18)) = f-1(16) = 4 & f -1(g-1(27)) = f-1(25) = 5.

Partial Order: A relation 'R' on a Set 'A' is said to be  partial order if
1) R is reflexive ie (a,a) belongs R, for every a that belongs to A
2) R is anti-symmetric ie,for every a,b that belongs to A and


3) R is transitive ie, for every a, b,c for every a,ab



Let us take a example

If A= {1,2,3}
R1={ }
It is not reflexive hence it is not in partial or post order
R2={(1,1),(2,2),(3,3)}
It is reflexive, antisymmetric and transitive hence it is in partial order relation
R3={(1,1),(2,2),(3,3),(1,2)(2,1)}
It is reflexive but it is not anti symmetric hence it is not in partial  order relation
R4={(1,1),(2,2),(3,3),(1,3),(2,3)}
It is reflexive, anti symmetric and transitive, hence it is  in partial order relation

R5={(1,1),(1,2),(2,3),(1,3)}
R6=A X A={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}

R6 is reflexive, it is symmetric as you can see elements (1,2) and (2,1).
 As it is symmetric it cannot be anti symmetric. Therefore partial order does not hold.