Example-1 :- Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R′ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation.
Solution :-
The school is boys school and no student of the school can be sister of any student of the school.
Hence, R = φ, showing that R is the empty relation. It is also obvious that the difference between heights
of any two students of the school has to be less than 3 meters. This shows that R′ = A × A is the universal relation.
Example-2 :- Let T be the set of all triangles in a plane with R a relation in T given by R = {(T₁, T₂) : T₁ is congruent to T₂}. Show that R is an equivalence relation.
Solution :-Every triangle is congruent to itself. So, R is reflexive. If (T₁, T₂) ∈ R ⇒ T₁ is congruent to T₂ ⇒ T₂ is congruent to T₁ ⇒ (T₂, T₁) ∈ R. So, R is symmetric. If (T₁, T₂), (T₂, T₃) ∈ R ⇒ T₁ is congruent to T₂ and T₂ is congruent to T₃ ⇒ T₁ is congruent to T₃ ⇒ (T₁, T₃) ∈ R. So, R is transitive. So, R is an equivalence relation.
Example-3 :- Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is perpendicular to L₂}. Show that R is symmetric but neither reflexive nor transitive.
Solution :-
A line L₁ can not be perpendicular to itself, i.e., (L₁, L₁) ∉ R. So, R is not reflexive.
If (L₁, L₂) ∈ R ⇒ L₁ is perpendicular to L₂ ⇒ L₂ is perpendicular to L₁ ⇒ (L₂, L₁) ∈ R. So, R is symmetric.
If L₁ is perpendicular to L₂ and L₂ is perpendicular to L₃, then L₁ can never be perpendicular to L₃.
In fact, L₁ is parallel to L₃, i.e., (L₁, L₂) ∈ R, (L₂, L₃) ∈ R but (L₁, L₃) ∉ R. So, R is not transitive.
Example-4 :- Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Solution :-
Here, {(1, 1), (2, 2), (3, 3)} ∈ R. So, R is reflexive.
Here, (1, 2) ∈ R but (2, 1) ∉ R. So, R is not symmetric.
Similarly, (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R. So, R is not transitive.
Example-5 :- Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.
Solution :-Here, 2 divides (a – a) for all a ∈ Z. So, R is reflexive. If (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, So R is symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b – c) is even. So, (a – c) is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z.
Example-6 :- Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.
Solution :-
Given that any element a in A, both a and a must be either odd or even, so that (a, a) ∈ R.
Here, (a, b) ∈ R ⇒ both a and b must be either odd or even ⇒ (b, a) ∈ R.
Similarly, (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be either even or odd simultaneously ⇒ (a, c) ∈ R.
So, R is an equivalence relation.
Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements of this subset are odd.
Similarly, all the elements of the subset {2, 4, 6} are related to each other, as all of them are even.
Also, no element of the subset {1, 3, 5, 7} can be related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd,
while elements of {2, 4, 6} are even.
Example-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 :-
Here, Two different students of the class can have not same roll number. So, f is one-one.
We can assume without any loss of generality that roll numbers of students are from 1 to 50.
This implies that 51 in N is not roll number of any student of the class,
so that 51 can not be image of any element of X under f. So, f is not onto.
Example-8 :- Show that the function f : N → N, given by f(x) = 2x, is one-one but not onto.
Solution :-For f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂. So, f is one-one. For 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1. So, f is not onto.
Example-9 :- Prove that the function f : R → R, given by f(x) = 2x, is one-one and onto.
Solution :-
Here f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂. So, f is one-one.
Also, given any real number y in R, there exists y/2 in R such that f(y/2) = 2 . (y/2) = y.
So, f is onto.
Example-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 :-
Here, f(1) = f(2) = 1. So, f is not one-one.
As given any y ∈ N, y ≠ 1, we can choose x as y + 1 such that f(y + 1) = y + 1 – 1 = y.
Also for 1 ∈ N, we have f(1) = 1. So, f is onto.
Example-11 :- Show that the function f : R → R, defined as f(x) = x₂, is neither one-one nor onto.
Solution :-Here, f(– 1) = 1 = f(1).So, f is not oneone. Also, the element – 2 in the co-domain R is not image of any element x in the domain R. So, f is not onto.
Example-12 :- Show that f : N → N, given by
is both one-one and onto.
Suppose f(x₁) = f(x₂). Note that if x₁ is odd and x₂ is even, then we will have x₁ + 1 = x₂ – 1,
i.e., x₂ – x₁ = 2 which is impossible. Similarly, the possibility of x₁ being even and x₂ being odd can also be ruled out,
using the similar argument. Therefore, both x₁ and x₂ must be either odd or even. Suppose both x₁ and x₂ are odd.
Then f(x₁) = f(x₂) ⇒ x₁ + 1 = x₂ + 1 ⇒ x₁ = x₂. Similarly, if both x₁ and x₂ are even,
then also f(x₁) = f(x₂) ⇒ x₁ – 1 = x₂ – 1 ⇒ x₁ = x₂. Thus, f is one-one.
Also, any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number 2r in
the co-domain N is the image of 2r – 1 in the domain N. Thus, f is onto.
Example-13 :- Show that an onto function f : {1, 2, 3} → {1, 2, 3} is always one-one.
Solution :-
Suppose that f is not one-one. Then there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.
Also, the image of 3 under f can be only one element. Therefore, the range set can have at the most two elements of
the co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one.
Example-14 :- Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto.
Solution :-
Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.
Hence, f has to be onto.
Example-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 and g(5) = g(9) = 11. Find gof.
Solution :-Here, gof(2) = g(f(2)) = g(3) = 7, gof(3) = g(f(3)) = g(4) = 7, gof(4) = g(f(4)) = g(5) = 11 and gof(5) = g(5) = 11.
Example-16 :- Find gof and fog, if f : R → R and g : R → R are given by f(x) = cos x and g(x) = 3x². Show that gof ≠ fog.
Solution :-
Here, gof(x) = g(f(x)) = g(cos x) = 3 (cos x)² = 3 cos²x.
Similarly, fog(x) = f(g(x)) = f(3x²) = cos(3x²).
Note that 3cos²x ≠ cos3x², for x = 0. Hence, gof ≠ fog.
Example-17 :- Show that if f:R-{7/5} → R-{3/5} is defined by f(x) = (3x + 4)/(5x - 7) and g:R-{3/5} → R-{7/5} is defined by g(x) = (7x + 4)/(5x - 3) then fog = IA and gof = IB, where A = R-{3/5}, IA b = R-{7/5}; (x) = x, ∀ x ∈ A, IB (x) = x, ∀ x ∈ B are called identity functions on sets A and B, respectively
Solution :-Thus, gof(x) = x, ∀x ∈ B and fog(x) = x, ∀x ∈ A, which implies that gof = IB and fog = IA.
Example-18 :- Show that if f : A → B and g : B → C are one-one, then gof : A → C is also one-one.
Solution :-
Suppose gof(x₁) = gof(x₂)
⇒ g(f(x₁)) = g(f(x₂))
⇒ f(x₁) = f(x₂), as g is one-one
⇒ x₁ =x₂, as f is one-one Hence, gof is one-one.
Example-19 :- Show that if f : A → B and g : B → C are onto, then gof : A → C is also onto.
Solution :-Given an arbitrary element z ∈ C, there exists a pre-image y of z under g such that g(y) = z, since g is onto. Further, for y ∈ B, there exists an element x in A with f(x) = y, since f is onto. Therefore, gof(x) = g(f(x)) = g(y) = z, showing that gof is onto.
Example-20 :-
Solution :-
Consider f : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6} defined as f(x) = x, ∀x and
g : {1, 2, 3, 4, 5, 6} → {1, 2, 3, 4, 5, 6} as g(x) = x,
for x = 1, 2, 3, 4 and g(5) = g(6) = 5. Then, gof(x) = x ∀x, which shows that gof is one-one.
But g is clearly not one-one
Example-21 :- Are f and g both necessarily onto, if gof is onto?
Solution :-
Consider f : {1, 2, 3, 4} → {1, 2, 3, 4} and
g : {1, 2, 3, 4} → {1, 2, 3}
defined as f(1) = 1,
f(2) = 2,
f(3) = f(4) = 3,
g(1) = 1, g(2) = 2 and
g(3) = g(4) = 3.
It can be seen that gof is onto but f is not onto.
Example-22 :- Let f : {1, 2, 3} → {a, b, c} be one-one and onto function given by f(1) = a, f(2) = b and f(3) = c. Show that there exists a function g : {a, b, c} → {1, 2, 3} such that gof = Iₓ and fog = Iᵧ, where, X = {1, 2, 3} and Y = {a, b, c}.
Solution :-
Consider g : {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2 and g(c) = 3.
It is easy to verify that the composite gof = Iₓ is the identity function on X and
the composite fog = Iᵧ is the identity function on Y.
Example-23 :- Let f : N → Y be a function defined as f(x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.
Solution :-
Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3, for some x in the domain N.
This shows that x = (y-3)/4. Define g : Y → N by g(Y) = (y-3)/4 ).
Now, gof(x) = g(f(x)) = g(4x + 3) = (4x+3-3)/4 = x and fog(y) = f(g(y)) = f{(y-3)/4} = 4(y-3)/(4) + 3 = y - 3 + 3 = y
This shows that gof = IN and fog = Iᵧ, which implies that f is invertible and g is the inverse of f.
Example-24 :- Let Y = {n² : n ∈ N} ⊂ N. Consider f : N → Y as f(n) = n². Show that f is invertible. Find the inverse of f.
Solution :-
An arbitrary element y in Y is of the form n², for some n ∈ N. This implies that n = √y .
This gives a function g : Y → N, defined by g(y) = √y .
Now, gof(n) = g(n²) = √n² = n and fog(y) = f(√y ) = (√y)² = y, which shows that gof = IN and fog = Iᵧ.
Hence, f is invertible with f -1 = g.
Example-25 :- Let f : N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f : N → S, where, S is the range of f, is invertible. Find the inverse of f.
Solution :-Let y be an arbitrary element of range f. Then y = 4x² + 12x + 15, for some x in N, which implies that y = (2x + 3)² + 6. This givesHence, gof =IN and fog =IS. This implies that f is invertible with f-1 = g.
Example-26 :- Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z, ∀x, y and z in N. Show that ho(gof) = (hog) of.
Solution :-
We have ho(gof) (x) = h(gof (x)) = h(g(f(x))) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = sin (6x + 4) . x ∈ N
Also, ((hog)of ) (x) = ( hog) (f(x)) = (hog) (2x) = h(g(2x)) = h(3(2x) + 4) = h(6x + 4) = sin (6x + 4), ∀x ∈ N.
This shows that ho(gof) = (hog)of.
Example-27 :- Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f(1) = a, f(2) = b, f(3) = c, g(a) = apple, g(b) = ball and g(c) = cat. Show that f, g and gof are invertible. Find out f-1, g-1 and (gof)-1 and show that (gof)-1 = f-1og-1.
Solution :-
Note that by definition, f and g are bijective functions.
Let f-1: {a, b, c} → (1, 2, 3} and
g-1 : {apple, ball, cat} → {a, b, c} be defined as
f-1{a} = 1,
f-1{b} = 2,
f-1{c} = 3,
g-1{apple} = a,
g-1{ball} = b and
g-1{cat} = c.
It is easy to verify that
f-1of = I{1, 2, 3},
f o f-1 = I{a, b, c},
g-1og = I{a, b, c} and
go g-1 = ID, where, D = {apple, ball, cat}.
Now, gof : {1, 2, 3} → {apple, ball, cat} is given by
gof(1) = apple,
gof(2) = ball,
gof(3) = cat.
We can define (gof)-1 : {apple, ball, cat} → {1, 2, 3} by
(gof)-1 (apple) = 1,
(gof)-1 (ball) = 2 and
(gof)-1 (cat) = 3.
It is easy to see that (gof)-1 o (gof) = I{1, 2, 3} and
(gof) o (gof)-1 = ID.
Thus, we have seen that f, g and gof are invertible.
Now, f-1og-1 (apple)
= f-1(g-1(apple))
= f-1(a) = 1
= (gof)-1 (apple) f-11og-1 (ball)
= f -1(g-1(ball))
= f -1(b) = 2
= (gof)-1 (ball) and f-1og-1 (cat)
= f-1(g-1(cat))
= f-1(c) = 3
= (gof)-1(cat). Hence (gof)-1 = f-1og-1.
The above result is true in general situation also.
Example-28 :- Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f-1, if it exists.
(a) f = {(1, 1), (2, 2), (3, 3)}
(b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1, 3), (3, 2), (2, 1)}
(a) It is easy to see that f is one-one and onto,
so that f is invertible with the inverse f-1 of f given by f-1 = {(1, 1), (2, 2), (3, 3)} = f.
(b) Since f(2) = f(3) = 1, f is not one-one, so that f is not invertible.
(c) It is easy to see that f is one-one and onto, so that f is invertible with f-1 = {(3, 1), (2, 3), (1, 2)}.
Example-29 :- Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R* of nonzero real numbers.
Solution :-
+ : R × R → R is given by (a, b) → a + b
– : R × R → R is given by (a, b) → a – b
× : R × R → R is given by (a, b) → ab
Since ‘+’, ‘–’ and ‘×’ are functions, they are binary operations on R.
But ÷: R × R → R, given by (a, b) → a/b , is not a function and hence not a binary operation,
as for b = 0, a/b is not defined.
Example-30 :- Show that subtraction and division are not binary operations on N.
Solution :-
– : N × N → N, given by (a, b) → a – b, is not binary operation,
as the image of (3, 5) under ‘–’ is 3 – 5 = – 2 ∉ N.
Similarly, ÷ : N × N → N, given by (a, b) → a ÷ b is not a binary operation,
as the image of (3, 5) under ÷ is 3 ÷ 5 = 3/5 ∉ N.
Example-31 :- Show that ∗ : R × R → R given by (a, b) → a + 4b² is a binary operation.
Solution :-
Since ∗ carries each pair (a, b) to a unique element a + 4b² in R, ∗ is a binary operation on R.
Example-32 :- Let P be the set of all subsets of a given set X. Show that ∪ : P × P → P given by (A, B) → A ∪ B and ∩ : P × P → P given by (A, B) → A ∩ B are binary operations on the set P.
Solution :-
Since union operation ∪ carries each pair (A, B) in P × P to a unique element A ∪ B in P, ∪ is binary operation on P.
Similarly, the intersection operation ∩ carries each pair (A, B) in P × P to a unique element A ∩ B in P,
∩ is a binary operation on P.
Example-33 :- Show that the ∨ : R × R → R given by (a, b) → max {a, b} and the ∧ : R × R → R given by (a, b) → min {a, b} are binary operations.
Solution :-
Since ∨ carries each pair (a, b) in R × R to a unique element namely maximum of a and b lying in R, ∨ is a binary operation.
Using the similar argument, one can say that ∧ is also a binary operation.
Example-34 :- Show that + : R × R → R and × : R × R → R are commutative binary operations, but – : R × R → R and ÷ : R∗ × R∗ → R∗ are not commutative.
Solution :-
Since a + b = b + a and a × b = b × a, ∀a, b ∈ R, ‘+’ and ‘×’ are commutative binary operation.
However, ‘–’ is not commutative, since 3 – 4 ≠ 4 – 3.
Similarly, 3 ÷ 4 ≠ 4 ÷ 3 shows that ‘÷’ is not commutative.
Example-35 :- Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not commutative.
Solution :-
Since 3 ∗ 4 = 3 + 8 = 11 and
4 ∗ 3 = 4 + 6 = 10,
showing that the operation ∗ is not commutative.
Example-36 :- Show that addition and multiplication are associative binary operation on R. But subtraction is not associative on R. Division is not associative on R∗.
Solution :-
Addition and multiplication are associative,
since (a + b) + c = a + (b + c) and
(a × b) × c = a × (b × c) ∀ a, b, c ∈ R.
However, subtraction and division are not associative, as
(8 – 5) – 3 ≠ 8 – (5 – 3) and
(8 ÷ 5) ÷ 3 ≠ 8 ÷ (5 ÷ 3).
Example-37 :- Show that ∗ : R × R → R given by a ∗ b → a + 2b is not associative.
Solution :-
The operation ∗ is not associative,
since (8 ∗ 5) ∗ 3 = (8 + 10) ∗ 3 = (8 + 10) + 6 = 24,
while 8 ∗ (5 ∗ 3) = 8 ∗ (5 + 6) = 8 ∗ 11 = 8 + 22 = 30.
Example-38 :- Show that zero is the identity for addition on R and 1 is the identity for multiplication on R. But there is no identity element for the operations – : R × R → R and ÷ : R∗ × R∗ → R∗.
Solution :-
a + 0 = 0 + a = a and a × 1 = a = 1 × a, ∀a ∈ R implies that 0 and 1 are identity elements
for the operations ‘+’ and ‘×’ respectively. Further, there is no element e in R with a – e = e – a,
∀a. Similarly, we can not find any element e in R∗ such that a ÷ e = e ÷ a, ∀a in R∗.
Hence, ‘–’ and ‘÷’ do not have identity element.
Example-39 :- Show that – a is the inverse of a for the addition operation ‘+’ on R and 1/a is the inverse of a ≠ 0 for the multiplication operation ‘×’ on R.
Solution :-
As a + (– a) = a – a = 0 and (– a) + a = 0, – a is the inverse of a for addition.
Similarly, for a ≠ 0, a × 1/a = 1 = 1/a × a implies that 1/a is the inverse of a for multiplication.
Example-40 :- Show that – a is not the inverse of a ∈ N for the addition operation + on N and 1/a is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1.
Solution :-
Since – a ∉ N, – a can not be inverse of a for addition operation on N,
although – a satisfies a + (– a) = 0 = (– a) + a.
Similarly, for a ≠ 1 in N, 1/a ∉ N, which implies that other than 1 no element of
N has inverse for multiplication operation on N.
Example-41 :- Show that, If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof) = (hog)of.
Solution :-
Given that ho(gof) (x) = h(gof(x)) = h(g(f(x))), ∀x in X and
(hog) of (x) = hog(f (x)) = h(g(f(x))), ∀x in X.
Hence, ho(gof) = (hog)of.
Example-42 :- Show that 2 Let f : X → Y and g : Y → Z be two invertible functions. Then gof is also invertible with (gof)-1 = f-1og-1.
Solution :-
To show that gof is invertible with (gof)-1 = f-1og-1,
it is enough to show that (f-1og-1)o(gof) = Iₓ and (gof)o(f-1og-1) = IZ.
Now, (f-1og-1)o(gof) = (( f-1og-1) og) of, by above example = (f-1o(g-1og)) of,
by above example, 1 = (f-1 oIY) of, by definition of g-1 = Iₓ.
Similarly, it can be shown that (gof )o(f-1og-1) = IZ.
