Types of Functions

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One-to-one function or one–one function

A function f: A→B where A,B are two non-empty sets is called a one-to-one function if no two distinct elements of A have the same image in B. That is, f: A→B is a one-to-one (or one–one) function if and only if (i f f).

x1, x2∈A and x1≠ x2 then f(x1) ≠f(x2)    or

if f(x1) = f(x2) x1= x2

Example 1

f  = {(a,x), (b,y), (c,z)} is a one–one function.

Example 2

Let f: R→R be defined by f(x) = x2. Is f one–one? Consider f(1) = (1)2                     = 1              f(-1) = (-1)2 = 1              f(x1) = f(x2) = 1 But x1≠ x2       1 ≠-1 So f is not a one–one function. A one–one function or one-to-one function is also called an injection.

Onto function

A function f: A→B, A,B are non-empty sets, is called an onto function if f(A)=B. That is f is onto if every element of the codomain B is the image of at least one element of the domain A. f: A→B is onto if and only if (iff) for every y ∈B there exists at least one x∈ A such that f(x) = y.

Example 3

Let f: {a, b, c} →{1,2,3} such that f(a) = 3, f(b) = 2, f(c) = 1. f is an onto function, since each element of {1,2,3} is an image of an element of {a, b, c}.

Example 4

Let f: {2, 4, 7} →{p, q, r} such that f(2) = q, f(4) = r, f(7)= r f is not onto as p is not the image of any element of {2, 4, 7}

Example 5

Let   f: N→ {-1, 1} defined by f(n) = 1 if f is odd,       = -1 if n is even. f is onto.

Example 6

Let f: R→R be defined by f(x) = 3x-5. Show that f is onto.

Solution:

Let y = f(x) = 3x - 5       y = 3x - 5 y + 5 = 3x   For every y ∈R there is an x ∈R

such that

An onto function is also called a surjection.

One–one and onto functions

A function f: A→B, A,B are non-empty, is called a one–one and onto function if it is both one–one and onto. This type of function is also called a bijection.

Example 7

If f   = { (p,1), (q,2), (r,3) } f is one–one and onto Since f (p) = 1           f(q) = 2           f(r) = 3 For every element of {1,2,3} is the image of element of {p,q,r}. So f is onto. Also, f is one–one since two distinct elements of {p,q,r} have two distinct images in {1,2,3}.

Example 8

Let f: R →R be defined by f(x) = 2x + 3 f is a one–one and onto function. Consider      -1≠ 1                 f(-1) ≠f(1)      as 2 (-1) +3 ≠2 * 1+3                 -2+3 ≠2+3                      1 ≠5        Let y = f(x) = 2x + 3                  y - 3 = 2x

So for every y ∈R there exists an x ∈R such that

So f is onto.

Example 9

Let f: R→R be defined by   f (x) = x2 - 1 Consider    f (2) = 22 -1                       = 4-1                       = 3.                 f(-2) = (-2)2-1                       = 4-1                       = 3                  f(2) = f(-2)               But 2 ≠-2 f is not one–one. So f is not a bijection.

Try these questions

1. State if the following functions are One–One         a. f1(x) = x2          f1: R→ R         b. f2(x) = -3x        f2: R→ R         Solutions:         a. f1 : R →R f1(x) = x2 f1is not One–One since     f1(-2) = (-2)2             = 4      f1(2) = (2)2             = 4     f1(-2) = f1(2)             = 4 but -2 ≠2       b. f2: R →R  f2(x) = -3x    f2(x1) ≠ f(x2) -3x1 ≠-3x2     x1 ≠x2 So f2 is one–one.     2. State if the following functions are onto         a. g1(x) = 2x3      g1: R→ R          b. g3: Z→ Z defined by g3 (x) = x-1         Solutions:         a. g1: R→ R   g1(x) = 2x3 Let y = g1 (x) = 2x3      y = 2x3    y/2 = x3       g1 is onto.       b.  g3: Z→ Z      g3(x) = x-1  g3is an onto function since g3 (Z) = Z. Or for every x ∈ Z there exists a y ∈ Z such that g(x) = y.     3. State if the following functions are bijections. (One–One and onto functions).         a. f1 = {(a,-1), (b,-2), (c,-3), (d, -4)}         b. f2 : R→ R defined by f2(x) = 4x-1         Solutions:          a. f1 = {(a,-1), (b,-2), (c,-3), (d,(-4)}     f1 is One–One since      -1 ≠-2    f1(a) ≠f1(b)   a≠ b f1 is onto as every element of {a,b,c,d} has an image in {-1,-2,-3,-4}.    f1 is a bijection.       b. f2 : R→ R where f2(x) = 4x-1 f2 is One–One since     f2(x1) = f2(x2) On canceling like terms   x1 = x2 f2 is onto since for every element x∈R there exists an element y∈ R such that     f(x) = y. f2 is a bijection.     4. Find the inverse functions (if they exist) of the following.         a. f: R→ R f | x | = 2x2-1       b. f: R- {3} →R - {1} defined by f(x) =         Solutions:         a. f: R→ R f (x) = 2x2-1 f is not One–One as            f(-1) = 2 - 1                   = 1              f(1) = 2 - 1                    = 1 So         f(-1) = f(1) But -1 ≠1 f-1 does not exist.       b. f: R- {3} →R - {1} defined by f(x) =      f is One–One      f is onto       f-1 exists. let   y =f(x)= y = y(x-3) =  (x + 3) yx-3y= x+3 yx - x =  3 + 3 x(y-1) =  3(y+1)        x = x=f--1(y) = Or f--1(x) =