# Exercise 1.2 Solutions

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Exercise 1.2
1. Show that the function f : R* → R* defined by {f(x) = \dfrac{1}{x}} is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
To show that f is one-one
Consider two elements x_1, x_2 such that {f(x_1) = f(x_2)}
{⇒ \dfrac{1}{x_1} = \dfrac{1}{x_2}}
{x_1 = x_2}
f is one-one.
To show that f is onto
As R* is defined as {f(x) = \dfrac{1}{x}},
And also R* is defined as the set of non-zero real numbers.
For any element y ∈ R*, there exists a corresponding element \dfrac{1}{y}, in the domain R*
f is onto.
To check whether f is one-one when f is defined as f : N → R*
Consider two elements x_1, x_2N such that
f(x_1) = f(x_2)
{\Rightarrow \dfrac{1}{x_1} = \dfrac{1}{x_2}}
{x_1 = x_2}
f is one-one.
To check whether f is onto when f is defined as f : N → R*
To be onto, every element yR*, should have a corresponding mapping element \dfrac{1}{y} in the domain N
As the numbers in the form of \dfrac{1}{y} can not all be natural numbers, there exists some elements in the co-domain R* which are not mapped to any element in the domain N. In otherwords, some elements like 5, 11 in the co-domain R* will not be the images of any elements in the domain N, as {\dfrac{1}{5}, \dfrac{1}{11} ∉ \bf{N}}. In otherwords, all the elements in the co-domain R* are not mapped.
f is not onto.

2. Check the injectivity and surjectivity of the following functions:
i.
f : N → N given by {f(x) = x^2}
ii.
f : Z → Z given by {f(x) = x^2}
iii.
f : R → R given by {f(x) = x²}
iv.
f : N → N given by {f(x) = x^3}
v.
f : Z → Z given by {f(x) = x^3}
2.i Check the injectivity and surjectivity of f : N → N given by {f(x) = x²}
To check whether f is injective(one-one):
Consider two elements x_1, x_2N such that
{f(x_1) = f(x_2)}
{x_1^2 = x_2^2}
{x_1 = \pm x_2}
As both x_1, x_2N, we consider only {x_1 = x_2} and ignore {x_1 = -x_2} (as the set of natural numbers N does not contain any negative numbers)
{x_1 = x_2}
f : N → N is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element yN in the co-domain, there should exist an element xN in the domain such that
{x = \pm\sqrt{y}}
As we know, every natural number yN is not a perfect square.
So, there exists some elements i.e. natural numbers which are not perfect squares in the co-domain which are not images of any elements in the domain.
For example, \sqrt{2} = 1.414 but 1.414 ∉ N.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
f is not surjective(onto).
f is injective but not surjective
2.ii Check the injectivity and surjectivity of f : Z → Z given by {f(x) = x^2}
To check whether f is injective(one-one):
Consider two elements x_1, x_2Z such that
{f(x_1) = f(x_2)}
{x_1^2 = x_2^2}
{x_1 = \pm x_2}
As both x_1, x_2Z, we need to consider both {x_1 = x_2} and also {x_1 = -x_2} (as the set of integers Z contains both positive and negative numbers)
⇒ The image of x_1 is not unique/distinctive (In otherwords, two distinct elements x_1, x_2 have the same image)
f : Z → Z is not injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element yZ in the co-domain, there should exist an element xZ in the domain such that
{x = ±\sqrt{y}}
As we know, every positive integer yZ is not a perfect square. Also, all the negative integers are not squares of any integers.
⇒ There exists some elements in the co-domain i.e. positive integers which are not perfect squares as well as all the negative integers, which are not images of any element in the domain.
f is not surjective(onto).
f is neither injective nor surgective.
2.iii Check the injectivity and surjectivity of f : R → R given by {f(x) = x^2}
To check whether f is injective(one-one):
Consider two elements x_1, x_2R such that
{f(x_1) = f(x_2)}
{x_1^2 = x_2^2}
{x_1 = \pm x_2}
As both x_1, x_2R, we need to consider both {x_1 = x_2} and also {x_1 = -x_2} (as the set of real numbers R contains both positive and negative numbers)
⇒ The image of x_1 is not unique/distinctive (In otherwords, two distinct elements x_1 and x_2 in the domain have can have the same image in the co-domain)
f : RR is not injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element yR in the co-domain, there should exist an element xR in the domain such that
{x = \pm\sqrt{y}}
As we know, all the negative real numbers are not squares of any other real numbers.
⇒ There exists some elements i.e. negative real numbers in the co-domain which are not images of any element in the domain.
f is not sujective(onto).
f is neither injective nor surjective.
2.iv Check the injectivity and surjectivity of f : N → N given by {f(x) = x^3}
To check whether f is injective(one-one):
Consider two elements x_1, x_2N such that
{f(x_1) = f(x_2)}
{x_1^3 = x_2^3}
{x_1 = x_2}
f : N → N is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^3}
For the function f to be onto, for every element yN in the co-domain, there should exist an element xN in the domain such that
{x = \sqrt{y}}
As we know, every natural number yN is not a perfect cube.
So, there exists some elements i.e. natural numbers which are not perfect cubes in the co-domain which are not images of any natural number in the domain.
For example, \sqrt{2} ≅ 1.26 but 1.26 ∉ N.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
f is not surjective(onto).
f is injective but not surjective.
2.v Check the injectivity and surjectivity of f : Z → Z given by {f(x) = x^3}
To check whether f is injective(one-one):
Consider two elements x_1, x_2Z such that
{f(x_1) = f(x_2)}
{x_1^3 = x_2^3}
{x_1 = x_2}
f : Z → Z is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^3}
For the function f to be surjective or onto, for every element yZ in the co-domain, there should exist an element xZ in the domain such that
{x = \sqrt{y}}
As we know, every integer yZ is not a perfect cube.
So, there exists some elements i.e. natural numbers which are not perfect cubes in the co-domain which are not images of any natural number in the domain.
For example, \sqrt{2} ≅ 1.26 but 1.26 ∉ Z.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
f is not surjective(onto).
f is injective but not surjective.

3. Prove that the Greatest Integer Function f : R → R, given by {f(x) = [x]}, is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
To check whether f is one-one:
The function f : R → R is defined as {f(x) = [x]}
For instance,
f(2) =  = 2
f(2.34) = [2.34] = 2
f(2.999) = [2.999] = 2
f(2.578344) = [2.578344] = 2
So, the same element 2 is image of many different elements in the domain.
⇒ Every integer element in the co-domain R is an image of multiple elements in the domain R.
f is not one-one.
To check whether f is onto:
The function f : R → R is defined as {f(x) = [x]}
This means that every element in the domain is mapped to an integer in the co-domain.
⇒ The elements in the co-domain which are not integers are not images of any element in the domain.
For instance, 1.23 is not an image of any element in the domain.
f is not onto.

4. Show that the Modulus Function f : R → R, given by {f(x) = |x|}, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is –x, if x is negative.
To check whether f is one-one:
The function f : R → R is defined as {f(x) = |x|}
Consider two elements x_1, x_2R in the domain R such that
{f(x_1) = f(x_2)}
{|x_1| = |x_2|}
x_1 = \pm x_2
⇒ Two different elements, one positive and one negative but equal in magnitude have the same image in the co-domain R.
f is not one-one.
To check whether f is onto:
The function f : R → R is defined as {f(x) = |x|}
As we know |x| is always positive
⇒ Every element xR is mapped to a positive real number yR
⇒ There are few negative numbers yR in the co-domain, which are not images of any element in the domain R
f is not onto.

5. Show that the Signum Function f : R → R, given by {f = \begin{cases}1, & x \gt 0\\0, & x = 0\\-1, & x \lt 0\end{cases}} is neither one-one nor onto.
To check whether f is one-one:
In the given function f, all the positive real numbers in the domain have the same image 1. Similarly all the negative numbers in the domain have the same image -1 in the co-domain.
⇒ Multiples elements in the domain have the same image in the domain.
For instance,
f(1) = 1
f(9) = 1
f(55) = 1
So, the distinct elements 1, 10, 55 in the domain have the same image 1 in the co-domain.
f is not one-one.
To check whether f is onto:
Elements other than -1, 0 and 1 in the co-domain are not images of any element in the domain.
For example, 10 is not image of any element.
f is not onto.

6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Observing A, every element in A has a unique/distinct image in B.
Given that A = {1, 2, 3} and B = {4, 5, 6, 7}
The function f is defined as f = {(1, 4), (2, 5), (3, 6)}
f(1) = 4, f(2) = 5 and f(3) = 6
This can be diagramatically represented as follows:
From the above, it is clear that every element in A has a unique image in B.
f is one-one.

7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
i.
f : R → R defined by {f(x) = 3 – 4x}
ii.
f : R → R defined by {f(x) = 1 + x^2}
To check whether f is one-one where f : R → R defined by {f(x) = 3 - 4x}
Consider two elements x_1, x_2R in the domain such that {f(x_1) = f(x_2)}
{3 - 4x_1 = 3 - 4x_2}
{x_1 = x_2}
f is one-one.
To check whether f is onto where f : R → R defined by {f(x) = 3 - 4x}
For every element yR should be an image of atleast one element x in the domain such that
{y = 3 - 4x}
{4x = 3 - y}
{⇒ x = \dfrac{3 - y}{4}}
⇒ Every element yR in the co-domain will be an image of an the element {\dfrac{3 - y}{4}} in the domain R
f is onto.
∴ As f is both one-one and onto, f is bijective.
To check whether f is one-one where f : R → R defined by {f(x) = 1 + x^2}
Consider two elements x_1, x_2R in the domain such that {f(x_1) = f(x_2)}
{1 + x_1^2 = 1 + x_2^2}
{x_1^2 = x_2^2}
{x_1 = \pm x_2}
⇒ Two distinct elements have the same image.
f is not one-one.
To check whether f is onto where f : R → R defined by {f(x) = 1 + x^2}
For every element yR should be an image of atleast one element x in the domain such that
{y = 1 + x^2}
As we know, x^2 is always positive. Also, 1 is a positive number.
So, the sum of two positive numbers i.e. 1 and x^2 is always positive.
y is always positive and can not be negative.
As we know, the co-domain R contains both positive and negative real numebers.
⇒ The negative numbers in the co-domain can not be images of any elements in the domain.
In otherwords, if you consider a random number -7 in the co-domain, there exists no element x in the domain R such that f(x) = -7.
f is not onto.
∴ As f is neither one-one nor onto, f is not bijective.

8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
To check whether f is one-one
In this function, the domain is A × B and the co-domain is B × A
Also note that every element in the domain A × B will be in the form {(a, b)} i.e. it not be a single element but it will be an ordered pair.
Now, the function f is defined as {f(a, b) = (b, a)}
Let (a_1, b_1), (a_2, b_2)A × B be two elements in the domain such that
{f(a_1, b_1) = f(a_2, b_2)}
{(b_1, a_1) = (b_2, a_2)} (∵ f is defined as {f(a, b) = (b, a)})
f is one-one.
To check whether f is onto.
As f is defined as {f(a, b) = (b, a)}, it can be interpreted that every element (b, a)B × A in the co-domain will be an image of the element (a, b)A × B in the domain.
f is onto.
As f is both one-one and onto, f will be bijective.

9. Let f : N → N be defined by \begin{cases}\dfrac{n + 1}{2} & ,\ if\ n\ is\ odd\\\dfrac{n}{2} & ,\ if\ n\ is\ even\end{cases} for all n ∈ N. State whether the function f is bijective. Justify your answer.
To check whether f is one-one
Let k be an odd natural number.
Then {k + 1} will be an even natural number.
Now {f(k) = \dfrac{k + 1}{2}} (∵ k is odd, we’ve to use {f(n) = \dfrac{n + 1}{2}})
Now {f(k + 1) = \dfrac{k + 1}{2}} (∵ {k + 1} is even, we’ve to use {f(n) = \dfrac{n}{2}})
⇒ Two different consecutive natural numbers have the same image.
f is not one-one
To check whether f is onto
To be onto, every natural number yN in the co-domain should be an image of an element in the domain N.
When x is odd {y = f(x) = \dfrac{x + 1}{2}}
{2y - 1 = x}
As we know, for any value of yN, {2y - 1} is always odd and also {2y - 1}N
⇒ Every yN in the co-domain will be an image of an element in the domain.
Now, when x is even {y = f(x) = \dfrac{x}{2}}
{2y = x}
As we know, for any value of yN, 2y is always even and also 2yN
⇒ Every yN in the co-domain will be an image of an element in the domain.
Thus we see that in all the cases, every element yN in the co-domain will be an image of an element in the domain. (to be specific every yN will be an image of two elements in the domain)
In otherwords, all the elements in the co-domain N will have an image in the domain
f is onto(surjective) but not one-one(injective).
f is not bijective.
Note: For short answer/MCQ, as soon as you discover that f is not one-one, you can conclude that f is not bijective. In otherwords, you don’t have to check whether f is onto. This might save some time.

10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by {f(x) = \left(\dfrac{x - 2}{x - 3}\right)}. Is f one-one and onto? Justify your answer.
The function f : A → B is given as
{f(x) = \left(\dfrac{x - 2}{x - 3}\right)}
{⇒ f(x) = \left(\dfrac{x - 3 + 1}{x - 3}\right)}
{⇒ f(x) = \left(\dfrac{x - 3}{x - 3} + \dfrac{1}{x - 3}\right)}
{⇒ f(x) = 1 + \dfrac{1}{x - 3}}
To Check whether f is one-one:
Let x_1, x_2A in the domain such that
{f(x_1) = f(x_2)}
{⇒ 1 + \dfrac{1}{x_1 - 3} = 1 + \dfrac{1}{x_2 - 3}}
{⇒ \dfrac{1}{x_1 - 3} = \dfrac{1}{x_2 - 3}}
{x_1 - 3 = x_2 - 3}
{x_1 = x_2}
f is one-one.
To Check whether f is onto.
For f to be onto, for every element yB in the co-domain should be a image of an element xA in the domain such that
{y = f(x) = 1 + \dfrac{1}{x - 3}}
{⇒ \dfrac{1}{x - 3} = y - 1}
{⇒ x - 3 = \dfrac{1}{y - 1}}
{⇒ x = 3 + \dfrac{1}{y - 1}}
The above expression is valid for all real values of y, except 1 (∵ {y - 1 = 0} when when {y = 0} and division by 0 is not defined)
It is given in the problem that 1 ∉ B,
⇒ Every element yB in the co-domain is an image of an element xA
f is onto.

11. Let f : R → R be defined as {f(x) = x^4}. Choose the correct answer.
A.
f is one-one onto
B.
f is many-one onto
C.
f is one-one but not onto
D.
f is neither one-one nor onto.
To Check whether f is one-one:
Consider two elements x_1, x_2R such that
{f(x_1) = f(x_2)}
{x_1^4 = x_2^4}
{x_1^2 = x_2^2}
{x_1 = \pm x_2}
⇒ Two different elements in the domain have the same image in the co-domain.
f is not one-one.
To Check whether f is onto:
For f to be onto, every element yR in the co-domain should be an image of some element xR in the domain, such that
{y = f(x) = x^4}
As we know, x^4 is always positive.
However, as the function f is defined as f : R → R, the co-domain, which is set of real numbers has both positive and negative numbers, will have only the positive numbers as the images of elements in the domain. In otherwords, the negative numbers will not be images of any elements in the domain.
f is not onto.
∴ As f is neither one-one nor onto, D is the correct answer.
12. Let f : R → R be defined as {f(x) = 3x}. Choose the correct answer.
A.
f is one-one onto
B.
f is many-one onto
C.
f is one-one but not onto
D.
f is neither one-one nor onto.
To Check whether f is one-one:
Consider two elements x_1, x_2R such that
{f(x_1) = f(x_2)}
{3x_1 = 3x_2}
{x_1 = x_2}
f is not one-one.
To Check whether f is onto:
For f to be onto, every element yR in the co-domain should be an image of some element xR in the domain, such that
{y = f(x) = 3x}
⇒ x = \dfrac{y}{3}
This mapping will be true for all values of y.
⇒ Every element y in the co-domain will be an image of an element x in the domain.
f is onto.
∴ As f is both one-one and onto, A is the correct answer.