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

   g₁(x) = 2x³ g₁: R→ R

   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)}

   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

   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) =