TOPICS
Unit-1(Examples)

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 examples is both one-one and onto.

Solution :-
  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 :-
  examples
  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 gives
  examples
  Hence, 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)}

Solution :-
  (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.
    
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