Function – Definition
A function or mapping (Defined as f: X→Y) is a relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets). X is called Domain and Y is called Codomain of function ‘f’.
Function ‘f’ is a relation on X and Y s.t for each x ∈X, there exists a unique y ∈ Y such that (x,y) ∈ R. x is called pre-image and y is called image of function f.
A function can be one to one, many to one (not one to many). A function f: A→B is said to be invertible if there exists a function g: B→A
Injective / One-to-one function
A function f: A→B is injective or one-to-one function if for every b ∈ B, there exists at most one a ∈ A such that f(s) = t.
This means a function f is injective if a1 ≠ a2 implies f(a1) ≠ f(a2).
Problem:
Prove that a function f: R→R defined by f(x) = 2x – 3 is a bijective function.
Explanation: We have to prove this function is both injective and surjective.
If f(x1) = f(x2), then 2x1 – 3 = 2x2 – 3 and it implies that x1 = x2.
Hence, f is injective.
Here, 2x – 3= y
So, x = (y+5)/3 which belongs to R and f(x) = y.
Hence, f is surjective.
Since f is both surjective and injective, we can say f is bijective.
Composition of Functions
Two functions f: A→B and g: B→C can be composed to give a composition g o f. This is a
function from A to C defined by (gof)(x) = g(f(x))
Example
Let f(x) = x + 2 and g(x) = 2x, find ( f o g)(x) and ( g o f)(x)
Solution
(f o g)(x) = f (g(x)) = f(2x) = 2x+2
(g o f)(x) = g (f(x)) = g(x+2) = 2(x+2)=2x+4
Hence, (f o g)(x) ≠ (g o f)(x)
Some Facts about Composition
* If f and g are one-to-one then the function (g o f) is also one-to-one.
* If f and g are onto then the function (g o f) is also onto.
* Composition always holds associative property but does not hold commutative property.
A function or mapping (Defined as f: X→Y) is a relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets). X is called Domain and Y is called Codomain of function ‘f’.
Function ‘f’ is a relation on X and Y s.t for each x ∈X, there exists a unique y ∈ Y such that (x,y) ∈ R. x is called pre-image and y is called image of function f.
A function can be one to one, many to one (not one to many). A function f: A→B is said to be invertible if there exists a function g: B→A
Injective / One-to-one function
A function f: A→B is injective or one-to-one function if for every b ∈ B, there exists at most one a ∈ A such that f(s) = t.
This means a function f is injective if a1 ≠ a2 implies f(a1) ≠ f(a2).
Example
1. f: N →N, f(x) = 5x is injective.
2. f: Z+→Z+, f(x) = x2 is injective.
3. f: N→N, f(x) = x 2 is not injective as (-x)2 = x2
Surjective / Onto function
A function f: A →B is surjective (onto) if the image of f equals its range. Equivalently, for
every b ∈ B, there exists some a ∈ A such that f(a) = b. This means that for any y in B,
there exists some x in A such that y = f(x).
Example
1. f : Z+→Z+, f(x) = x2 is surjective.
2. f : N→N, f(x) = x2 is not injective as (-x)2 = x2
Bijective / One-to-one Correspondent
A function f: A →B is bijective or one-to-one correspondent if and only if f is both injective
and surjective.
1. f: N →N, f(x) = 5x is injective.
2. f: Z+→Z+, f(x) = x2 is injective.
3. f: N→N, f(x) = x 2 is not injective as (-x)2 = x2
Surjective / Onto function
A function f: A →B is surjective (onto) if the image of f equals its range. Equivalently, for
every b ∈ B, there exists some a ∈ A such that f(a) = b. This means that for any y in B,
there exists some x in A such that y = f(x).
Example
1. f : Z+→Z+, f(x) = x2 is surjective.
2. f : N→N, f(x) = x2 is not injective as (-x)2 = x2
Bijective / One-to-one Correspondent
A function f: A →B is bijective or one-to-one correspondent if and only if f is both injective
and surjective.
Problem:
Prove that a function f: R→R defined by f(x) = 2x – 3 is a bijective function.
Explanation: We have to prove this function is both injective and surjective.
If f(x1) = f(x2), then 2x1 – 3 = 2x2 – 3 and it implies that x1 = x2.
Hence, f is injective.
Here, 2x – 3= y
So, x = (y+5)/3 which belongs to R and f(x) = y.
Hence, f is surjective.
Since f is both surjective and injective, we can say f is bijective.
Composition of Functions
Two functions f: A→B and g: B→C can be composed to give a composition g o f. This is a
function from A to C defined by (gof)(x) = g(f(x))
Example
Let f(x) = x + 2 and g(x) = 2x, find ( f o g)(x) and ( g o f)(x)
Solution
(f o g)(x) = f (g(x)) = f(2x) = 2x+2
(g o f)(x) = g (f(x)) = g(x+2) = 2(x+2)=2x+4
Hence, (f o g)(x) ≠ (g o f)(x)
Some Facts about Composition
* If f and g are one-to-one then the function (g o f) is also one-to-one.
* If f and g are onto then the function (g o f) is also onto.
* Composition always holds associative property but does not hold commutative property.
No comments:
Post a Comment