# 11th Class Mathematics Relations Notes - Mathematics Olympiads - Relation and Function

Notes - Mathematics Olympiads - Relation and Function

Category : 11th Class

Relation and Function

Let $A=\{1,\,2,\,3,4,\}$, $B=\{2,\,3\}$

$A\times B=\{1,\,2,3,\,4,\}\times \{2,3\}=\{(1,2),(2,2),(3,2),(4,2),(1,3),(2,3),(3,3),(4,3)\}$

Let we choose an arbitrary set:

$R=[(1,2),(2,2),(1,3),(4,3)]$

Then R is said to be the relation between a set A to B.

Definition

Relation R is the subset of the Cartesian Product$A\times B$. It is represented as $R=\{(x,y):x\in A\,$ and $y\in B\}$ {the 2nd element in the ordered pair (x, y) is the image of 1st element}

Sometimes, it is said that a relation on the set A means the all members / elements of the relation

R be the elements / members of $A\text{ }\times \text{ }A$.

e.g.      Let $A=\{1,\,2,\,3\}$ and a relation R is defined as $R=\{(x,y):x<y$ where $x,y\in A\}$

Sol.     $\because$$\mathbf{A=\{1,}\,\mathbf{2,}\,\mathbf{3\}}$

$A\times A=\{(1,1),(2,2),(3,3),(2,1),(3,1),(1,2),(3,2),(1,3),(2,3)\}$

$\because \,\,\,\,R=\,\,\,\because x<y$

$\because \,\,\,\,R=\{(x,y):x<y,and\,x,y\in A\}=\{(1,2),(2,3),(1,3)$

Note: Let a set A has m elements and set B has n elements. Then $n(A\times B)$ be $m\times n$elements so, total number of relation from A to B or between A and B be${{2}^{m\times n}}$.

• A relation can be represented algebraically either by Roster method or set builder method.

Types of Relation

• Void Relation: A relation R from A to B is a null set, then R is said to be void or empty relation.

• Universal Relation: A relation on a set A is said to be universal relation, if each element of A is related to or associated with every element of A.

• Identity Relation: A relation ${{I}_{x}}\{(x,x):x\in A\}$ on a set A is said to be identity relation on A.

• Reflexive relation: A relation Ron the set A is said to be the reflexive relation. If each and every element of set A is associated to itself. Hence, R is reflexive iff $(a,\,a)\in R\,\,\forall \,\,a\in A$.

e.g.       $\Rightarrow \,\,A=\{1,\,2,\,3,\,4\}$

$R=\{(1,\,3),(1,1),(2,3),(3,2),(2,2),(3,1),(3,3),(4,4)\}$is a reflexive relation on f.

Sol.     Yes, because each and every element of A is related to itself in R.

• Symmetric relation: A relation R on a set A is said to be symmetric relation iff. $(x,y)\in R\Rightarrow (y,x)\in R\,\,\forall \,\,x,y\in A$

i.e. $x\,R\,y\Rightarrow y\,R\,x\,\,\forall \,\,x,y\in A$

$\because$ xRy is read as x is R-related to y.

• Anti-symmetric relation: A relation which is not symmetric is said to be anti-symmetric relation.

• Transitive relation: Let A be any non-empty set. A relation R on set A is said to be transitive relation R iff $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R\,\,\forall \,\,x,y,z\in R.$

i.e.       $xRy$ and $yRz\Rightarrow xRz\,\,\forall \,\,x,\,y,\,z\in R$.

1. Let $\mathbf{A=\{1,}\,\mathbf{2,}\,\mathbf{3,}\,\mathbf{4\}}$

$A\times A=\{(1,,1),(2,1),(3,1),(4,1),(1,2),(2,2),(3,2),(4,2),(1,3),(2,3),(3,3),(4,3),(1,4),(2,4),(3,4),(4,4),$

${{R}_{1}}=(1,1),(2,2),(3,2),(2,3),(3,3),(4,4)$

${{R}_{2}}=(2,2),(1,3),(3,3),(3,1),(1,1)$

${{R}_{3}}=(1,1),(2,2),(3,4),(3,3),(4,4)$

State about${{\mathbf{R}}_{\mathbf{1}}}$, ${{\mathbf{R}}_{\mathbf{2}}}$ and${{\mathbf{R}}_{\mathbf{3}}}$. Are they reflexive, symmetric, anti-symmetric or transitive relations?

Sol.     ${{R}_{1}}$ is symmetric as well as transitive relation for ${{R}_{2}}$.

${{R}_{2}}$is not reflexive because $(4,4)\in {{R}_{2}}$ But ${{R}_{2}}$ is symmetric as well as transitive.

Now,

${{R}_{3}}$is reflexive as well as anti-symmetric because $(3,4)\in {{R}_{3}}$ but $(4,3)\notin {{R}_{3}}$.

• Equivalence Relation

A relation R on a set A is said to be an equivalence relation on A iff

(i)        It is reflexive i.e. $(x,y)\in R\,\,\forall \,\,x\in R$

(ii)       It is symmetric i.e. $(x,y)\in R$

$\Rightarrow (x,y)\in R\,\,\forall \,\,x,y\in R$

(ii)       It is $(x,y)\in R(y,z)\,\in R$ then $(x,z)\in R$

$\forall \,\,x,y,z\in R$

e.g.      Let $A=\{x,y,z\}$ and R is defined as $R=\{(x,y),(y,y),(z,z),(x,y)(y,z)\}$

Sol.     Yes, because R is satisfied reflexive symmetric as well transitivity condition of relation R on A.

• Calculus: It is the mathematics of change and motion.

• Function: Concept of function play a very vital role in mathematics either pure mathematics or applied mathematics. In the secondary school classes, we will learn about real value function only. Function is defined as the following ways:

(a)       Function as an operation.

(b)       Function as a relation.

(c)        Function as a mapping.

Type of Function

• Function as an operator

• Function: Function is an operator in which we study the relation between independent and dependent real variable/input or output variable.

 Input $f$ Output

e.g.       $y=f(x)$

$y=\text{f}\left( x \right)={{x}^{2}}$ As we know that function contain in the pure mathematics. We cherish both faces of mathematics, the pure as a beautiful retreat from reality and applied an ardent hope for life.

• Function as a mapping: Let A and B be two non-empty sets. Each element in the set A associated with unique element in set B is said to be function or mapping from A to B.

i.e.       $f:A\to B.$ e.g. $y=f\left( x \right)$

x is said to be independent variable and y is said to be dependent variable.

All element of set A is said to be domain of f and all elements of set B is said to be co-domain off as well as all image of each element of set A in B, is said to be the range of the function /.

$y=f(x)\Rightarrow f(A)\subseteq B$. e.g. $y=f(A)\le B$.

• Function as a relation

• Relation: Relation on a set is the Cartesian product of two non-empty sets. Function is a relation but relation may be or may not be function because function has some salient features.

e.g.      $f:A\to B$

(i)        For every $x\in A$, $y\in B$ such that $y=\text{f}(x)$. i.e. each element of set A has image in B. But there may be an element in B which is not the image of element of A.

(ii)       Same element of A cannot be associated to distinct element of B. i.e. the image of each element of set A has unique. But the distinct element of A may be associated to same element of B.

• Basically Function be Two Type

(i) Into function and onto function (surjective function)

(ii) Into function be two types.

1.a One-one into function (Injective function)

1.b Many-one into function

2.a One-one onto function (bijective function)

2.b Many-one onto function

• Some Standard Functions
1. Constant function
2. Square function
3. Identity function
4. Linear function
5. Cube function
6. Reciprocal function
7. Step function
8. Modulus function
9. Logarithmic function
10. Exponential function
11. Greatest integer function
12. Polynomial function
13. Polynomial ratio function
14. Trigonometric function
15. Inverse function.

Function

• Type of Function

1. One-one function (Injective function): A function $\text{f:}\,\text{A}\to \text{B}$ is said to one-one function. If the distinct image.

if${{x}_{1}}\ne {{x}_{2}}\Leftrightarrow f\left( {{x}_{1}} \right)\ne f\left( {{x}_{2}} \right)\,\,\forall {{x}_{1}},{{x}_{2}}\in R$.

1. Many One function: A function $\text{f}\,\,\text{: }\,\text{A}\to \text{B}$ is said to be many one iff it is not one - one.

e.g. $f\,\,\text{: }\,\text{A}\to \text{B}$is defined as $\text{y=}(x)={{x}_{2}}$ is a many one function

3.         Onto function or surjective function: • A function f: A$\to$B is said to be onto function (surjective function) iff each element of B is the image of all element of A i.e. range of the function is equal to the co-domain of the function.

e.g.      $\because x\in A$ then $f\left( x \right)=y\in B$

iff $f\left( A \right)=B$

i.e. range of f= codomain off. This is onto function.

• One-one onto function or (bijective function): A function $f\,\,:\,\,\,A\,\to B$is said to be one-one onto or bijective function iff distinct element of A has distinct images in B as well as range of / is equal to the domain of f e. f ((A)=B

e.g.      A function $f\,\,:\,\,\,A\,\to B$ is defined as $\text{y=}f(x)=x$ is 1-1 onto function #### Other Topics

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