concept of relation and functions class 12
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The concept of relation is used in relating two objects or quantities with each other. Suppose two sets are considered, the relationship between them will be established if there is a connection between the elements of two or more non-empty sets.
Mathematically, “a relation R from a set A to a set B is a subset of the cartesian product A × B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A × B”.
Types of Relations
A relation R from A to A is also stated as a relation on A, and it can be said that the relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two extreme relations. Below are the definitions of types of relations:
Empty Relation
If no element of A is related to any element of A, i.e. R = φ ⊂ A × A, then the relation R in a set A is called empty relation.
Universal Relation
If each element of A is related to every element of A, i.e. R = A × A, then the relation R in set A is said to be universal relation.
Both the empty relation and the universal relation are some times called trivial relations.
A relation R in a set A is called-
Reflexive- if (a, a) ∈ R, for every a ∈ A,
Symmetric- if (a1, a2) ∈ R implies that (a2, a1) ∈ R , for all a1, a2∈ A,
Transitive- if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R for all a1, a2, a3 ∈ A.
Equivalence Relation- A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
Functions
A function is a relationship which explains that there should be only one output for each input. It is a special kind of relation(a set of ordered pairs) which obeys a rule, i.e. every y-value should be connected to only one y-value.
Mathematically, “a relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B”.
In other words, a function f is a relation from a set A to set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element. Also, A and B are two non-empty sets.
Types of Functions
One to one Function: A function f : X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct, i.e., for every x1 , x2 ∈ X, f(x1 ) = f(x2 ) implies x1 = x2 . Otherwise, f is called many-one.
One-one function
many-one function
Onto Function: A function f: X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f(x) = y.
Onto function
One-one and Onto Function: A function f: X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto.
One-one and Onto function
Composition of 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)), ∀ x ∈ A
Invertible Functions
A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f–1.
An important note is that, if f is invertible, then f must be one-one and onto and conversely if f is one-one and onto, then f must be invertible.
Binary Operations
A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a, b) by a ∗ b.
Example Problems
Example 1: Show that subtraction and division are not binary operations on R.
Solution: N × N → N, given by (a, b) → a – b, is not binary operation, as the image of (2, 5) under ‘–’ is 2 – 5 = – 3 ∉ N.
Similarly, ÷: N × N → N, given by (a, b) → a ÷ b is not a binary operation, as the image of (2, 5) under ÷ is 2 ÷ 5 = 2/5 ∉ N.
Example 2: Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof.
Solution: From the given, we have:
gof(2) = g (f(2)) = g (3) = 7
gof (3) = g (f(3)) = g (4) = 7
gof(4) = g (f(4)) = g (5) = 11
gof(5) = g (5) = 11
Example 3: Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.
Solution: R is reflexive, as 2 divides (a – a) for all a ∈ Z.
Further, if (a, b) ∈ R, then 2 divides a – b.
Therefore, 2 divides b – a.
Hence, (b, a) ∈ R, which shows that R is symmetric.
Similarly, if (a, b) ∈ R and (b, c) ∈ R, then (a – b) and (b – c) are divisible by 2.
Now, a – c = (a – b) + (b – c) is even. (from the above statements)
From this,
(a – c) is divisible by 2.
This shows that R is transitive.
Thus, R is an equivalence relation in Z.
Answer:
here is the answer:-
The concept of relation is used in relating two objects or quantities with each other. Suppose two sets are considered, the relationship between them will be established if there is a connection between the elements of two or more non-empty sets.
Mathematically, “a relation R from a set A to a set B is a subset of the cartesian product A × B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A × B”.
Types of Relations
A relation R from A to A is also stated as a relation on A, and it can be said that the relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two extreme relations. Below are the definitions of types of relations:
Empty Relation
If no element of A is related to any element of A, i.e. R = φ ⊂ A × A, then the relation R in a set A is called empty relation.
Universal Relation
If each element of A is related to every element of A, i.e. R = A × A, then the relation R in set A is said to be universal relation.
Both the empty relation and the universal relation are some times called trivial relations.
A relation R in a set A is called-
Reflexive- if (a, a) ∈ R, for every a ∈ A,
Symmetric- if (a1, a2) ∈ R implies that (a2, a1) ∈ R , for all a1, a2∈ A,
Transitive- if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R for all a1, a2, a3 ∈ A.
Equivalence Relation- A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
Functions
A function is a relationship which explains that there should be only one output for each input. It is a special kind of relation(a set of ordered pairs) which obeys a rule, i.e. every y-value should be connected to only one y-value.
Mathematically, “a relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B”.
In other words, a function f is a relation from a set A to set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element. Also, A and B are two non-empty sets.
Types of Functions
One to one Function: A function f : X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct, i.e., for every x1 , x2 ∈ X, f(x1 ) = f(x2 ) implies x1 = x2 . Otherwise, f is called many-one.
One-one function
many-one function
Onto Function: A function f: X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f(x) = y.
Onto function
One-one and Onto Function: A function f: X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto.
One-one and Onto function
Composition of 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)), ∀ x ∈ A
Invertible Functions
A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f–1.
An important note is that, if f is invertible, then f must be one-one and onto and conversely if f is one-one and onto, then f must be invertible.
Binary Operations
A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a, b) by a ∗ b.
Example Problems
Example 1: Show that subtraction and division are not binary operations on R.
Solution: N × N → N, given by (a, b) → a – b, is not binary operation, as the image of (2, 5) under ‘–’ is 2 – 5 = – 3 ∉ N.
Similarly, ÷: N × N → N, given by (a, b) → a ÷ b is not a binary operation, as the image of (2, 5) under ÷ is 2 ÷ 5 = 2/5 ∉ N.
Example 2: Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof.
Solution: From the given, we have:
gof(2) = g (f(2)) = g (3) = 7
gof (3) = g (f(3)) = g (4) = 7
gof(4) = g (f(4)) = g (5) = 11
gof(5) = g (5) = 11
Example 3: Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.
Solution: R is reflexive, as 2 divides (a – a) for all a ∈ Z.
Further, if (a, b) ∈ R, then 2 divides a – b.
Therefore, 2 divides b – a.
Hence, (b, a) ∈ R, which shows that R is symmetric.
Similarly, if (a, b) ∈ R and (b, c) ∈ R, then (a – b) and (b – c) are divisible by 2.
Now, a – c = (a – b) + (b – c) is even. (from the above statements)
From this,
(a – c) is divisible by 2.
This shows that R is transitive.
Thus, R is an equivalence relation in Z.
Step-by-step explanation: