__Class 12 chapter 1__

__Relations and functions__

*Basic concepts of topic relations and functions class XII chapter 1 of mathematics. Equivalence relations, different types of functions, composition and inverse of functions.*

**Introduction**

**In class 11 we have studied about
Cartesian product of two sets, relations, functions, domain, range and
co-domains. Now in this chapter we have studied about the different types of
relations, different types of functions, composition of functions and
invertible functions.**

**Relations**

**A relation R from set A to set B is the subset of Cartesian product A x B.**

**In Cartesian product A x B number of relations equal to the number of subsets.**

**Domain**

**Set of all first elements of the ordered pairs in the relation R is called its domain.**

**Range**

**Set of all second elements of the ordered pairs in the relation R is called its range.**

**In any ordered pair second element is also called the image of first element.**

**Types of relations**

**i) Empty Relation:**

**A relation R in set A is called empty
relation, if no element of A is related to any element of A, i.e., R = Ï† ****⊂ A x A**

**Example: Let A = {1, 2, 3} and R be a relation on the set A defined as R = {(a, b) : a + b = 10; a, b ****∈ A}**

**Here R is empty relation. R = Î¦**

**ii) Identity Relation**

**A relation R in set A is called identity relation, if every element of A is related to itself, i.e., R = {a, a) ****⊂ A x A**

**A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.**

**iii) Universal Relation :**

**A relation R in a set A is called universal relation, if each
element of A is related to every element of A, i.e., R = A x A**

**iv) Reflexive relation**

**A relation R in a set A is called reflexive, if (a, a) ∈ R, for every a ∈ A**

**v) Symmetric Relation**

**A relation R in a set A is called symmetric, if (a, b) ∈R implies that (b, a) ∈ R, for all a, b ∈ A.**

**vi) Transitive Relation**

**A relation R in a set A is called transitive, if (a, b) ∈R and (b, c) ∈R, implies that (a, c) ∈ R, for all a, b, c**_{ }∈ A

**vii) Equivalence relation:**

**A relation R in set A is said to be equivalence relation if it
is reflexive, symmetric and transitive
relation.**

**Viii) Equivalence Relations:**

**Let R be an equivalence relation on a set A and let a ****∈ A. Then we define equivalence class of a as **

**[a] ={b ****∈ A : b is related to a} = {b ****∈ A : (b, a) ****∈ R}**

**Note : Both empty relation and universal relations are sometimes called equivalence relation of trivial relation.**

**Functions **

**One-One functions (or injective
functions):**

**A function f : A →B is said to be one –one or injective function if
different elements of set A has different images in set B**

**Mathematically: **

**Case 1: For all a, b ****∈ A, we have: If f(****a) = f(b****) ****⇒ ****a**_{ }= b

**Case 2 : For a, b ****∈ A, such that : If ****a ****≠ b**** ****⇒ ****f(a) ≠ f(b)**

**In the above figure different elements of set A have different
images in set B. So this function is called one-one function.**

**Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is = **^{4}P_{3} = 4! = 24.

**Generalization**

**Set A has m elements and the set B has n elements (n > m). Then the number of injective functions that can be defined from set A to set B is = **^{n}P_{m} = n!

**If n < m then ****Then the number of injective functions that can be defined from set A to set B is = 0**

**If a function has m elements and is defined from A to A then number of injective relations from A to A = n!**

**Many One functions:**

**If two or more elements of set A have same image in set B then it is called many one function.**

** If a function is not one-one then it is
called many one.**

**Here two different elements a and b
of set A has same image in set B. So it is not an one-one function. It is only
a many one function.**

**Onto function ( or surjective
function)**

**A function f : A →B is said to be onto (or
surjective) function, if every element of set B is the image of some element in
set A.**

**Let y ****∈ B, then for every element
y ∈ B, ∃ (there exist) an element x ∈ A
such that f(x) = y**

**In the above figure every element of set B is the image of some
element in set A so it is a onto function. It is neither many one nor one-one.**

**If n(A) = p, then number of bijective functions from set A to A are p!**

Because If n(A) = p and n(A) = n(B) then the number of bijective functions from A to B is p!

**Other definitions of onto functions are as follows**

**A function f : A →B is said to be onto (or
surjective) function, if every element of set B has pre - image in set A.**

**If range of the function is equal to the co-domain(or set B),
then it is called onto function.**

**Exercise 1.2**

**Q No. 9**

**\[f:N\rightarrow N,\; \; f(n)=\left\{\begin{matrix} \frac{n+1}{2}, if\; \; n\; \; is\; \; odd\\ \frac{n}{2}, if\; \; n \; \; is\; \; even \end{matrix}\right.\]**

**State whether the function f is bijective. Justify your answer.**

__One-One__

**\[Let\; (1,2)\epsilon N\; then\]\[f(1)=\frac{n+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\]\[f(2)=\frac{n}{2}=\frac{2}{2}=1\]\[f(1)=f(2)\: but\: \: 1\neq 2\]⇒ f is not one one function.**

__Onto__

**Case 1: If n = odd = 2q+1=4q+1\[f(n)=\frac{n+1}{2}\]\[f(4q+1)=\frac{4q+1+1}{2}=\frac{4q+2}{2}=2q+1\]⇒ 2q + 1 is also an odd function ∈ N**

**Case 2: If n = even = 2q = 4q, \[f(n)=\frac{n}{2}=\frac{4q}{2}=2q\]⇒ 2q is also an even function ****∈ N**

**From case 1 and case 2 we conclude that :**

***For all odd numbers in the domain we have odd numbers in the co-domain .**

***For all even numbers in the domain we have even numbers in the co-domain .**

**⇒ ****R**_{f} = Odd natural number and Even natural numbers.

**⇒ ****R**_{f} = All Natural Numbers = Co-domain

**⇒ f(x) is an onto function.**

**Now f(x) is an onto function but not one-one.**

**So f(x) is not a bijective function.**

****************************************************************************

**Composition of two functions**

**Let f : A →B and g
: B →C be two functions.
Then the composition of f and g, denoted by gof, is defined as the function gof
: A →C given by **

**gof(x) = g[f(x)], ****∀ (for all) x ****∈ A**

**Composition of two functions exists only if co-domain of first function
is equal to the domain of second function**

**In the above figure function f is
defined on the element of set A (i.e.
x), function g is defined on the elements of set B (i.e f(x) ) so
composition from set A to set C becomes gof or g[f(x)]**

**Bijective functions:**

**The functions which are one-one and
onto both are called bijective functions.**

**Invertible functions and existence of inverse**

**If a function f is one-one and onto
then it is an invertible function.**

**If a function is an invertible
function then inverse of f exist and is denoted by f**^{-1}.

**If f is invertible the f must be
one-one and onto, conversely if f is one-one and onto then f must be invertible. If f is invertible
then f**^{-1} exists.

**Let f(x) = y is an invertible
function then f**^{-1} exists and we can write f^{-1}(y) = x

**Example : f(x) = { (a, 1), (b, 2),
(c, 3), (d, 4)}, this can be written as**

**f(a) = 1, f(b) = 2, f(c) = 3,
f(d) = 4**

**If f is invertible then we can write**

**f**^{-1}(1) = a, f^{-1}(2) = b, f^{-1}(3) = c, f^{-1}(4) = d

**If f and g are two invertible
functions then their composition gof is also invertible.**

**Give an example of a relation which is**

**i) Symmetric but neither reflexive nor transitive**

**Ans R = { (1,2), (2,1)}**

**ii) Transitive but neither reflexive nor symmetric**

**Ans: R = { (1,1), (2,2), (1,2), (2,1), (3,4)}**

**iii) Reflexive and symmetric but not transitive.**

**Ans: R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}**

**iv) Reflexive and transitive but not symmetric.**

**Ans: R = {(1,1), (2,2), (1,2), (2,1), (2,3), (3,3)}**

**v) Symmetric and transitive but not reflexive**

**{(5,6), (6,5),(5,5) }**

** **__THANKS FOR YOUR VISIT__

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Very nice and helpful thanks

ReplyDeleteThank u sir . Useful

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