17.4 - Composition of Functions

In the previous section, we completed a discussion on one-one and onto functions. In this section, we will see Composition of functions.

Composition of functions

Composition of functions can be explained using an example. It can be written in 8 steps:
1. The display board of a textile shop reads: 14% discount on all items.
2. A person wants to buy a shirt. The marked price of the shirt is Rs 400/-. He wants to know whether he can afford it. He has to consider both the discount and the tax. The tax is to be paid at the rate of 8%.
3. For solving this problem, first we consider the discount.
• We know that:
If x is the marked price, then the cost after 14% discount will be 0.86x
• We can write this as a function: f(x) = 0.86x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the marked price.
    ♦ Output ‘f(x)’ is the cost after discount.
• In our present case, we can put x₁ = 400
So the cost after discount = f(x₁) = 0.86 × 400 = 344
4. Next we consider the tax.
• We know that:
If x is the cost of an item, the payment to be made considering 8% tax will be 1.08x
• We can write this as a new function: g(x) = 1.08x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the cost after discount.
    ♦ Output ‘g(x)’ is the payment to be made.
    ♦ In our present case, we can put x₂ = 344
So the payment = g(x₂) = 1.08 × 344 = 371.52
5. Note that, we did two separate calculations:
(i) In the first calculation, we found out the cost after discount.
(ii) In the second calculation, we found out the payment to be made after considering the tax.
6. The two calculations can be done together. It can be explained in 4 steps:
(i) Output of the first function is the input of the second function.
• Output of the first function is 0.86x. Here x is the marked price.
(ii) So input of the second function is 0.86x
• The second function is: g(x) = 1.08x
So we get: g(0.86x) = 1.08(0.86x) = 0.9288x
• Check:
0.9288 × 400 = 371.52
This is the same result that we obtained in (4)
(iii) Now we can write the general form:
• From the above steps, it is clear that:
    ♦ Output of f
    ♦ is the
    ♦ Input of g  
• Using symbols, this can be written in two ways:
    ♦ (g∘f)(x)
    ♦ g(f(x))
We can use any one of the above two ways. Both are read as "g of f of x"
(iv) Note that, the small circle between g and f should not be a filled circle. Then it would look like a dot. If it is a dot, then it means multiplication.
7. (g∘f)(x) is actually a function. It can be plotted just like any other function. The plot in our present case is the magenta line in fig.17.6 below:

Fig.17.6

An explanation of the graph can be written in 5 steps:
(i) Draw the three graphs:
    ♦ f(x) = 0.86x
    ♦ g(x) = 1.08x
    ♦ (g∘f)(x) = 0.9288x
(ii) We want the payment amount when the marked price is 400. So draw a vertical cyan dashed line through 400 on the x-axis.
• This vertical line meets f(x) at a point. Draw a horizontal cyan dashed line through this meeting point.
• This horizontal cyan dashed line meets the y-axis at 344. This is the price after discount.
(iii) We want the payment amount when the cost is 344. So draw a vertical brown dashed line through 344 on the x-axis.
• This vertical line meets g(x) at a point. Draw a horizontal brown dashed line through this point.
• This horizontal brown dashed line meets the y-axis at 371.52. This is the payment amount.
(iv) The horizontal cyan dashed line and vertical brown dashed line, together with the axes, form a square. Why?
(v) In practice, we need not plot f(x) and g(x). All we need is the plot of (g∘f)(x).
• This is because:
The vertical dashed line through 400 meets the magenta line at the white dot. The horizontal dashed line through this white dot meets the y-axis at the same 371.52.
8. (g∘f)(x) can be explained using set theory also. It can be written in 5 steps:
(i) We write set M which contains the marked prices of the various items in the shop. This is shown in fig.17.7(a) below:

Fig.17.7

 
(ii) Each element of M becomes input of the function f. The outputs are the costs after discount. These outputs become elements of set D. Let us see an example:
• 735 is an input for function f. We get: f(735) = 0.86 × 735 = 632.1
• So there is an arrow connecting 735 (in set M) and 632.1 (in set D).
• We can define the function f using sets M and D:
f: M→D defined as f(x) = 0.86x
(iii) Each element of D becomes input of the function g. The outputs are the payment amounts. These outputs become elements of set P. Let us see an example:
• 632.1 is an input for function g. We get: g(632.1) = 1.08 × 632.1 = 682.668
• So there is an arrow connecting 632.1 (in set D) and 682.668 (in set P).
• We can define the function g using sets D and P:
g: D→P defined as g(x) = 1.08x
(iv) Instead of applying two separate functions, we can apply the single function (g∘f). This is shown in fig.17.7(b) above. Let us see an example:
• (g∘f)(735) = 0.9288 × 735 = 682.668
We get the same result as above.
• We see an arrow directly connecting 725 (in set M) and 682.668 (in set P).
• We can define the function (g∘f) using sets M and P:
(g∘f): M→P defined as (g∘f)(x) = 0.9288x
• Note that, M is the domain of the first function and P is the codomain of the second function. So we can write:
The composite function connects the domain of the first function to the codomain of the second function.
(v) So it is clear that, by using composite functions, we can obtain the answers in a single step.    

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Solved example 17.15
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,
g(5) = g(9) = 11.
Find g∘f.
Solution:
1. We can name the given sets as shown below:
A = {2, 3, 4, 5}
B = {3, 4, 5, 9}
C = {7, 11, 15}
• So the functions can be indicated as:
f : A → B
g : B → C
2. The Venn diagram is shown in fig.17.8(a) below:

Fig.17.8

Let us see an example from the diagram.
• Given that, f(3) = 4,
So there is an arrow connecting 3 (in set A) and 4 (in set B).
• Given that, g(4) = 7,
So there is an arrow connecting 4 (in set B) and 7 (in set C).
3. By following the arrows in fig.a, we get the connection between elements in A and C. This is shown in fig.b above.
• So we can write:
(g∘f)(2) = 7
(g∘f)(3) = 7
(g∘f)(4) = 11
(g∘f)(5) = 11

Alternate method:
(g∘f) means, "output of f" is used as "input of g".
1. Given that, f(2) = 3
So output of f for the input 2 is 3
• This 3 is the input for g.
• Given that, g(3) = 7
So when the input for (g∘f) is 2, the output will be 7
We can write: (g∘f)(2) = 7
2. Given that, f(3) = 4
• Given that, g(4) = 7
So (g∘f)(3) = 7
3. Given that, f(4) = 5
• Given that, g(5) = 11
So (g∘f)(4) = 11
4. Given that, f(5) = 5
• Given that, g(5) = 11
So (g∘f)(5) = 11

Solved example 17.16
Find g∘f and f∘g, if f : R → R and g : R → R are given by f(x) = cos x and g(x) = 3x² . Show that g∘f ≠ f∘g.
Solution:
Part (i): Finding g∘f
1. g∘f means, "output of f" is used as "input of g".
2. Output of f is cos x.
So we can write:
g∘f = g(cos x) = 3(cos x)² = 3 cos²x

Part (ii): Finding f∘g
1. f∘g means, "output of g" is used as "input of f".
2. Output of g is 3x².
So we can write:
f∘g = f(3x²) = cos (3x²)

Part (iii): proving that, g∘f ≠ f∘g
1. We have: g∘f = 3 cos²x
Put x = 0. We get:
g∘f = 3 cos²0  = 3 × 1² = 3
2. We have f∘g = cos (3x²)
Put x = 0. We get:
f∘g = cos (3 × 0) = cos 0 = 1
3. So we can write: g∘f ≠ f∘g

Solved example 17.17
Show that,
• if $f: R- \left \{\frac{7}{5} \right\}~\to~R- \left \{\frac{3}{5} \right\}$ is defined by $f(x)~=~\frac{3x+4}{5x-7}$
• and $g: R- \left \{\frac{3}{5} \right\}~\to~R- \left \{\frac{7}{5} \right\}$ is defined by $f(x)~=~\frac{7x+4}{5x-3}$,
• then f∘g = I_A and g∘f = I_B,
• where
    ♦ $A = R- \left \{\frac{3}{5} \right\}$
    ♦ $B = R- \left \{\frac{7}{5} \right\}$
    ♦ $I_A (x) = x, \forall x \in A$
    ♦ $I_B (x) = x, \forall x \in B$
Solution:
Details about identity function can be seen here.

Part (i): To prove that (f∘g)(x) = I_A(x).
1. First we find (f∘g)
• (f∘g) means, "output of g" is used as "input of f".
Output of g is $\frac{7x+4}{5x-3}$.
So we get:
$$\begin{array}{ll}{}    &{(f \circ g)(x)}    & {~=~}    &{f \left(\frac{7x+4}{5x-3} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{3\left(\frac{7x+4}{5x-3} \right)+4}{5\left(\frac{7x+4}{5x-3} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+12 + 20x – 12}{35x + 20 -35x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (f∘g)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(f \circ g)(x)~=~f \left(\frac{7x+4}{5x-3} \right)$
• When x is $\frac{3}{5}$, the denominator will become zero. So x can be any real number other that $\frac{3}{5}$.
• That means, the domain is $R- \left \{\frac{3}{5} \right\}$
• But we are given that $A = R- \left \{\frac{3}{5} \right\}$.
So the domain is set A.
• We can write:
(f∘g)(x) is an identity function whose domain is A.
• Using symbols, we write this as:
(f∘g)(x) = I_A(x).

Part (ii): To prove that (g∘f)(x) = I_B(x)
1. First we find (g∘f)
• (g∘f) means, "output of f" is used as "input of g".
Output of f is $\frac{3x+4}{5x-7}$.
So we get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g \left(\frac{3x+4}{5x-7} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{7\left(\frac{3x+4}{5x-7} \right)+4}{5\left(\frac{3x+4}{5x-7} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+28 + 20x – 28}{15x + 20 -15x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (g∘f)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(g \circ f)(x)~=~g \left(\frac{3x+4}{5x-7} \right)$
• When x is $\frac{7}{5}$, the denominator will become zero. So x can be any real number other that $\frac{7}{5}$.
• That means, the domain is $R- \left \{\frac{7}{5} \right\}$
• But we are given that $B = R- \left \{\frac{7}{5} \right\}$.
So the domain is set B.
• We can write:
(g∘f)(x) is an identity function whose domain is B.
• Using symbols, we write this as:
(g∘f)(x) = I_B(x)

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In the next section, we will see a few more solved examples. We will also see some important properties of composite functions.

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17.3 - Prove A Function to be one-one or onto

In the previous section, we saw some different types of functions. We also saw solved examples demonstrating how to check whether a function is one-one or onto. In this section, we will see a few more solved examples.

Solved example 17.10
Show that the function
f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,
for every x > 2,
is onto but not one-one.
Solution:
To get a better understanding about the given function, it can be written as a piecewise function:

$$f(x) =
\begin{cases}
1,  & \text{if $x$ is 1} \\[1.5ex]
1,  & \text{if $x$ is 2} \\[1.5ex]
x-1 & \text{if $x$ > 2}
\end{cases}$$

1. f is a function from N to N.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
2. We take x from the set N. Set N is the set of natural numbers.
3. For each x value, we get the corresponding y value. This y value is also from set N.
4. Given that: f (1) = f (2) = 1
So we can write:
• the first ordered pair in f will be: (1, 1)
• the second ordered pair in f will be: (2, 1)
5. We see that, "1" is the image of both 1 and 2.
So f is not one-one.
6. Next, we will prove that, f is onto. For that, we need to prove that, all elements in the codomain (N) will be an image.
• We already saw that, "1" will be an image. So we need not consider it again.
• We must show the general case for the infinite number of natural numbers coming after "1"  
7. Consider any natural number y (other than 1) from the set N.
• Choose the input x such that x = y+1
• y is not equal to 1. So y can be 2, 3, 4, 5, . . .
• Then x will be always greater than 2. So the condition applicable is: f(x) = x-1  
• We get:
$\begin{array}{ll}{}    &{f(x)}    & {~=~}    &{x-1}    &{} \\
{\Rightarrow}    &{f(y+1)}    & {~=~}    &{y+1-1}    &{} \\
{\Rightarrow}    &{f(y+1)}    & {~=~}    &{y}    &{} \\
\end{array}$
8. We can write:
• For every y in the set N, we have a (y+1). If we input that (y+1) in f, we will get y as the output.
• That means, every element in N will be an image. So f is an onto function.

Solved example 17.11
Show that the function f : R → R,
defined as f (x) = x² , is neither one-one nor onto.
Solution:
1. Let us write some elements in f.
(i) When x = 0, f(x) = f(0) = 0² = 0
So we get the element (0,0)
(ii) When x = 1, f(x) = f(1) = 1² = 1
So we get the element (1,1)
(iii) When x = -1, f(x) = f(-1) = (-1)² = 1
So we get the element (-1,1)
2. From (ii) and (iii) of the above step, we see that:
"1" is the image of both 1 and -1.
• So f is not an one-one function.
3. This will become more clear from the graph in fig.17.5 below.

Fig.17.5

Mark the points -1 and 1 on the x axis. Draw vertical green dashed lines upwards. Those vertical lines will meet the graph at two points. Through those points, draw a horizontal green dashed line. This horizontal line will meet the y axis at a point. This point "1" on the y axis is the image of both -1 and 1.

4. Next we prove that f is not on to.
• Pick the point "-1" from the codomain. This "-1" cannot be the image of any number in the domain. This is because, we are squaring the input x. The square cannot be -ve.
• That means, all elements of the codomain are not images. So f is not an onto function.
5. This will become more clear from the graph in fig.17.5 above.
• We see that:
We can draw a horizontal dashed line through any -ve y value. But such a line will never meet the graph. So we will not be able to find an x value such that, it's image is a -ve value.

Solved example 17.12
Show that f : N → N, given by
$$f(x) =
\begin{cases}
x+1,  & \text{if $x$ is odd} \\[1.5ex]
x-1, & \text{if $x$ is even}
\end{cases}$$
is both one-one and onto.
Solution:
• f is a function from N to N.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• We take x from the set N. Set N is the set of natural numbers 1,2,3, . . .  up to infinity.
• For each x value, we get the corresponding y value. This y value is also from set N.

1. In an one-one function, if f(x₁) is to be equal to f(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is an one-one function.
2. In our present case, suppose that, f(x₁) is equal to f(x₂). Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{x_1 + 1}    & {~=~}    &{x_2 -1 ~~\color{green}{\text{- - - I}}}    &{} \\
{\Rightarrow}    &{x_2 - x_1}    & {~=~}    &{2}    &{} \\
\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we assume that x₁ is odd and x₂ is even.
   ♦ Since x₁ is odd, f(x) will be (x₁ + 1)
   ♦ Also, since x₂ is even, f(x) will be (x₂ - 1)

• The difference between an odd number and an even number cannot be 2.
• So we can write:
The condition of “x₁ is odd and x₂ is even” is not possible. 

3. Again, suppose that, f(x₁) is equal to f(x₂), assuming x₁ to be even and x₂ to be odd. Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{x_1 - 1}    & {~=~}    &{x_2 + 1 ~~\color{green}{\text{- - - I}}}    &{} \\
{\Rightarrow}    &{x_1 - x_2}    & {~=~}    &{2}    &{} \\
\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we assume that x₁ is even and x₂ is odd.
   ♦ Since x₁ is even, f(x) will be (x₁ - 1)
   ♦ Also, since x₂ is odd, f(x) will be (x₂ + 1)

• The difference between an odd number and an even number cannot be 2.
• So we can write:
The condition of “x₁ is even and x₂ is odd” is not possible.

4. Based on (2) and (3), we can write:
Either x₁ and x₂ both must be odd.
Or x₁ and x₂ both must be even.
5. Suppose that, both x₁ and x₂ are odd. Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{x_1 + 1}    & {~=~}    &{x_2 + 1}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$ 

6. Suppose that, both x₁ and x₂ are even. Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{x_1 - 1}    & {~=~}    &{x_2 - 1}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$

7. From (5) and (6), it is clear that:
If f(x₁) is equal to f(x₂), then it implies that x₁ = x₂.
• So f is an one-one function.   

8. Next, we will prove that, f is onto.
• Suppose that, the input x value is odd. We know that, any odd number can be represented as (2r+1)
• So we get:
f(x) = f(2r+1) = (2r +1 + 1) = 2r+2
• But 2r +2 is an even number.
• So we can write:
By suitably varying the value of r in the input odd number, we can obtain all even numbers as outputs.

9. Suppose that, the input x value is even. We know that, any even number can be represented as (2r)
• So we get:
f(x) = f(2r) = 2r - 1
• But 2r - 1 is an odd number.
• So we can write:
By suitably varying the value of r in the input even number, we can obtain all odd numbers as outputs.

10. From (8) and (9), it is clear that:
All numbers in the codomain N can become images. So f is an onto function.

Solved example 17.13
Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one.
Solution:
1. Given that, f is onto.
So one or more arrows will be converging on each and every element of the codomain {1,2,3}
2. We have to prove that, f is one-one. That is., only one arrow will be converging on each and every element of the codomain {1,2,3}
3. f is a function. So a single arrow will be diverging from each and every element of the domain {1,2,3}
4. Suppose that, f is not one-one. Then there are two possibilities.
(i) All three arrows diverging from  the domain, converge on a single element in the codomain.
(ii) Two arrows diverging from the domain, converge on a single element in the codomain.
• The third arrow from the domain, converges on one of the two remaining elements in the codomain.
5. Consider the possibility written in 4(i).
• Here, two elements in the codomain are left without any converging arrows.
• This contradicts with the statement that we wrote in (1). So thus possibility can be ruled out.
6. Consider the possibility written in 4(ii).
• Here, one element in the codomain is left without any converging arrows.
• This contradicts with the statement that we wrote in (1). So thus possibility can be ruled out.
7. We see that, both the possibilities that we wrote in (4), are ruled out. So f is an one-one function.

Solved example 17.14
Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto.
Solution:
1. Given that, f is one-one.
So, if any element in the codomain {1,2,3} is an image, then it will have only one arrow converging on it.
2. We have to prove that, there is an arrow converging on each and every element in the codomain {1,2,3}.
3. f is a function. So a single arrow will be diverging from each and every element of the domain {1,2,3}
4. Suppose that, f is not onto. Then there are two possibilities.
(i) All three arrows diverging from  the domain, converge on a single element in the codomain.
(ii) Two arrows diverging from the domain, converge on a single element in the codomain.
The third arrow from the domain, converges on one of the two remaining elements in the codomain.
5. Consider the possibility written in 4(i).
This contradicts with the statement that we wrote in (1). So thus possibility can be ruled out.
6. Consider the possibility written in 4(ii).
This contradicts with the statement that we wrote in (1). So thus possibility can be ruled out.
7. We see that, both the possibilities that we wrote in (4), are ruled out. So f is  an onto function.

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Link to a few more solved examples is given below:

Exercise 17.2

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In the next section, we will see composition of functions and invertible functions.

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17.2 - Types of Functions

In the previous section, we completed a discussion on relations. In this section, we will see functions.

We have seen the basics about functions in class 11. Details here. Now we will see different types of functions.

One-one function and many-one function

This can be explained in 2 steps:
1. Consider the function f₁ in fig.17.2(a) below:

Fig.17.2

• We see that:
f₁(1) = a, f₁(2) = b, f₁(3) = d, and f₁(4) = c.
• Note that, if an element in X₂ is an image, it has only one arrow converging on it.
• So each element in X₁ has a unique image in X₂.
• Such functions are called one-one functions.
• One-one functions are also called injective functions.
• We can write:
In one-one functions, [f(x₁) = f(x₂)] is possible only if x₁ = x₂.
• The function f₄ in fig.17.2(d) is also a one-one function.

2. Consider the function in fig.17.2(b) above:
• We see that:
f₂(1) = b, f₂(2) = b, f₂(3) = c, and f₂(4) = d
• Note that, the image b in X₂ has more than one arrows converging on it.
• So the elements 1 and 2 in X₁ do not have unique images in X₂. They have a common image, which is b.
• Such functions are called many-one functions.
• The function f₂ in fig.(b) is a many-one function.
• The function f₃ in fig.(c) is also a many-one function.

Onto function

This can be explained as follows:
• Consider the function f₃ in fig.17.2(c) above.
• We see that:
f₃(1) = a, f₃(2) = a, f₃(3) = b, and f₁(4) = c,
• There is not even a single element in X₃, which is not an image.
• Such functions are called onto functions.
• Onto functions are also called surjective functions.
• We can write:
In an onto function f: X⟶Y, the range will be same as set Y.
• We know that, set Y is called codomain. So we can write:
In an onto function, the range is same as the codomain. 
• The function f₄ in fig.17.2(d) is also an onto function.

One-one and onto function

This can be explained as follows:
• If a function is both one-one and onto, then it is called an one-one and onto function.
• Such functions are also called bijective functions.
• The function f₄ in fig.d above is an one-one and onto function.

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Now we will see some solved examples

Solved example 17.7
Let A be the set of all 50 students of Class X in a school. Let f : A → N be function defined by f (x) = roll number of the student x. Show that f is one-one but not onto.
Solution:
1. f is a function from A to N.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
2. We take x from the set A. Set A contains names of all 50 students in class X.
3. For each x value, we get the corresponding y value from set N. Set N is the set of natural numbers 1,2,3, . . .  up to infinity.
4. We can write:
• the first ordered pair in f will be: (name of first student, 1)
• the second ordered pair in f will be: (name of second student, 2)
• the third ordered pair in f will be: (name of third student, 3)
• so on . . .
5. We see that:
No two names can have the same roll number. That means, no two elements in A has the same image in N
• So f is an one-one function.
6. We also see that:
The elements coming after 50, are not images of any element in A. That means, there are some elements in N, which cannot become an image of any element in A. So f is not onto.
7. Note two points I and II:
Point I: This is related to one-one functions
(i) In an one-one function,
   ♦ the image of any element in the domain
   ♦ cannot be the
   ♦ image of any other element in the domain.
• Let x₁ and x₂ be any two elements in the domain. If it is an one-one function, f(x₁) has to be different from f(x₂).
• In an one-one function, if f(x₁) is to be equal to f(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is an one-one function.
(ii) To prove a function to be not one-one, we just need to pick a suitable element from the codomain and show that: It is the image of more than one elements in the domain.

[Recall that, to prove a statement to be true, we need to prove the general case. But to prove a statement to be false, it is sufficient to show an example using any convenient sample value]

Point II: This is related to onto functions
(i) To prove a function to be onto, we have to show that:
   ♦ each and every element of the codomain
   ♦ is an image.
(ii) To prove a function to be not onto, we just need to pick any suitable element of the codomain and show that: It is not an image.
In our present case, “51” is not an image.

Solved example 17.8
Show that the function f : N → N, given by f(x) = 2x, is one-one but not onto.
Solution:
• f is a function from N to N.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• We take x from the set N. Set N is the set of natural numbers 1,2,3, . . .  up to infinity.
• For each x value, we get the corresponding y value. This y value is also from set N.

1. In an one-one function, if f(x₁) is to be equal to f(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is an one-one function.
2. In our present case, suppose that, f(x₁) is equal to f(x₂). Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{2 x_1}    & {~=~}    &{2 x_2}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$
• So f is an one-one function.
3. This will become more clear from the graph in fig.17.3 below. Mark any natural number say 5, on the x axis. Draw a vertical green dashed line upwards. This vertical line will meet the graph at a point. Through that point, draw a horizontal green dashed line. This horizontal line will meet the y axis at 10. This 10 is the image of 5. There is only one possible image for 5, which is 10.

Fig.17.3

 
4. To prove a function to be not onto, we just need to pick any suitable element of the codomain and show that: It is not an image.
• In our present case, let us pick "7" from the codomain.
• If it is an image, then we can write an equation:
2x = 7
• The solution of this equation is not a natural number.
• So "7" cannot be the image of any element in the domain.
• Therefore f is not an onto function.
• This will become more clear from the graph in fig.17.3 above. Mark any odd natural number say 7, on the y axis. Draw a horizontal magenta dashed line towards the right. This horizontal line will not meet the graph at any point. That means, 7 is not the image of any number in N.

Solved example 17.9
Prove that the function f : R → R, given by f (x) = 2x, is one-one and onto.
Solution:
• f is a function from R to R.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• We take x from the set R. Set R is the set of real numbers.
• For each x value, we get the corresponding y value. This y value is also from set R.

1. In an one-one function, if f(x₁) is to be equal to f(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is an one-one function.
2. In our present case, suppose that, f(x₁) is equal to f(x₂). Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{2 x_1}    & {~=~}    &{2 x_2}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$
• So f is an one-one function.
3. This will become more clear from the graph in fig.17.4 below. Mark any number say $\sqrt{23} = 4.7958$, on the x axis. Draw a vertical green dashed line upwards. This vertical line will meet the graph at a point. Through that point, draw a horizontal green dashed line. This horizontal line will meet the y axis at a point. This point ($2 \sqrt{23}$) on the y axis is the image of $\sqrt{23}$. There is only one possible image for $\sqrt{23}$.

Fig.17.4

4. To prove a function to be onto, we have to show that:
   ♦ any element of the codomain
   ♦ is an image.
• In our present case, let "y" be any element of the codomain. Then we can write the equation:
2x = y
• "y" is an element of the codomain. The codomain is the set R. So "y" is a real number.
• Since "y" is a real number, we can surely find a real number "x" which is the solution of the equation 2x = y
• That means, we can pick any element "y" from the codomain. It will be an image.
• Therefore, f is an onto function.
• This will become more clear from the graph in fig.17.4 above. Mark any real number say -7, on the y axis. Draw a horizontal magenta dashed line towards the left. This horizontal line will meet the graph at a point. Through this point on the graph, draw a vertical magenta dashed line. This vertical line will meet the x axis at the point "-3.5". That means, there is a real number (-3.5) in R which has -7 as the image.

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In the next section, we will see a few more solved examples.

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