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