TOPICS
Exercise - 1.1

Question-1 :-  Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}

Solution :-
(i)   A = {1, 2, 3......13, 14}
      R = {(x, y): 3x − y = 0}
          3x - y = 0;
          3x = y;
          y = 3x (put the values of A in x)
      R = {(1, 3), (2, 6), (3, 9), (4, 12)} 
    ∴R is not reflexive, because (1, 1), (2, 2)....(14, 14) ∉ R. 
    ∴R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
    ∴R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0]
      Hence, R is neither reflexive, nor symmetric, nor transitive.    
       
 
(ii)  R = {(x, y): y = x + 5 and x < 4} 
      x = {1, 2, 3};
      y = {6, 7, 8}
      R = {(1, 6), (2, 7), (3, 8)}
      Here, (1, 1) ∉ R.
    ∴R is not reflexive. 
      (1, 6) ∈ R. But,(6, 1) ∉ R. 
    ∴R is not symmetric.
      Now, since there is no pair in R such that (x, y) and (y, z) ∈ R, then (x, z) ∉ R.
    ∴R is not transitive.
      Hence, R is neither reflexive, nor symmetric, nor transitive.
    
(iii) A = {1, 2, 3, 4, 5, 6}
      R = {(x, y): y is divisible by x} (means that y/x and remainder is 0).
      We know that any number (x) is divisible by itself. So, (x, x) ∈R 
    ∴R is reflexive.
      Now,(2, 4) ∈ R [as 4 is divisible by 2]. But, (4, 2) ∉ R. [as 2 is not divisible by 4]
    ∴R is not symmetric.
      Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. z is divisible by x.
   ⇒ (x, z) ∈R
    ∴R is transitive.
      Hence, R is reflexive and transitive but not symmetric.
       
(iv)  R = {(x, y): x − y is an integer}
      Now, for every x ∈ R, (x, x) ∈R as x − x = 0 is an integer.
    ∴R is reflexive.
      Now, for every x, y ∈ R  if (x, y) ∈ R, then x − y is an integer.
   ⇒ −(x − y) is also an integer.
   ⇒  (y − x) is an integer. (y, x) ∈ R
    ∴R is symmetric.
      Now, Let (x, y) and (y, z) ∈R, where x, y, z ∈ R   .
   ⇒ (x − y) and (y − z) are integers.
   ⇒ x − z = (x − y) + (y − z) is an integer. (x, z) ∈ R
    ∴R is transitive.
      Hence, R is reflexive, symmetric, and transitive.
        
(v)   Relation R in the set A of human beings in a town at a particular time
(a)   R = {(x, y): x and y work at the same place}
      (x, x) ∈ R
    ∴R is reflexive.
      If (x, y) ∈ R, then x and y work at the same place.
   ⇒ (y, x) ∈ R. y and x work at the same place.
    ∴R is symmetric.
      Now, let (x, y), (y, z) ∈ R
   ⇒ x and y work at the same place and y and z work at the same place.
   ⇒ x and z work at the same place.
   ⇒ (x, z) ∈R
    ∴R is transitive.
      Hence, R is reflexive, symmetric, and transitive.
    
(b)   R = {(x, y): x and y live in the same locality}
      (x, x) ∈ R as x and x is the same human being.
    ∴R is reflexive.
     If (x, y) ∈R, then x and y live in the same locality.
   ⇒ y and x live in the same locality.
   ⇒ (y, x) ∈ R
   ∴R is symmetric.
     Now, let (x, y) ∈ R and (y, z) ∈ R.
   ⇒ x and y live in the same locality and y and z live in the same locality.
   ⇒ x and z live in the same locality.
   ⇒ (x, z) ∈ R
   ∴R is transitive.
   Hence, R is reflexive, symmetric, and transitive.
    
(c)  R = {(x, y): x is exactly 7 cm taller than y} 
     Now, (x, x) ∉ R
     Since human being x cannot be taller than himself.
    ∴R is not reflexive. Now, let (x, y) ∈R.
   ⇒ x is exactly 7 cm taller than y.
     Then, y is not taller than x.
   ∴(y, x) ∉R
     Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
   ∴R is not symmetric.
     Now, Let (x, y), (y, z) ∈ R.
   ⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
   ⇒ x is exactly 14 cm taller than z .
   ∴(x, z) ∉R
   ∴R is not transitive.
     Hence, R is neither reflexive, nor symmetric, nor transitive.
    
(d)  R = {(x, y): x is the wife of y}
     Now, (x, x) ∉ R
     Since x cannot be the wife of herself.
   ∴R is not reflexive. Now, let (x, y) ∈ R
   ⇒ x is the wife of y.
     Clearly y is not the wife of x.
   ∴(y, x) ∉ R
     Indeed if x is the wife of y, then y is the husband of x.
   ∴R is not transitive. Let (x, y), (y, z) ∈ R
   ⇒ x is the wife of y and y is the wife of z.
     This case is not possible. Also, this does not imply that x is the wife of z.
   ∴(x, z) ∉ R
   ∴R is not transitive.
     Hence, R is neither reflexive, nor symmetric, nor transitive.
 
(e)  R = {(x, y): x is the father of y}
     Now, (x, x) ∉ R
     As x cannot be the father of himself.
   ∴R is not reflexive. 
     Now, let (x, y) ∈R.
   ⇒ x is the father of y.
   ⇒ y cannot be the father of y.
     So, y is the son or the daughter of y.
   ∴(y, x) ∉ R
   ∴R is not symmetric.
     Now, let (x, y) ∈ R and (y, z) ∈ R.
   ⇒ x is the father of y and y is the father of z.
   ⇒ x is not the father of z.
     So, x is the grandfather of z.
   ∴(x, z) ∉ R
   ∴R is not transitive.
     Hence, R is neither reflexive, nor symmetric, nor transitive.
 

Question-2 :-  Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Solution :-
    R = {(a, b): a ≤ b²}
    It can be observed that a is not equal to square of itself.  
  ∴R is not reflexive.
    Now, (1, 5) ∈ R as 1 < 5²
    But, 5 is not less than 1².
  ∴(5, 1) ∉ R
  ∴R is not symmetric.
    Now, (3, 2), (2, 1.5) ∈ R
    (as 3 < 2² = 4 and 2 < (1.5)² = 2.25)
    But, 3 > (1.5)² = 2.25
  ∴(3, 1.5) ∉ R
  ∴ R is not transitive.
    Hence, R is neither reflexive, nor symmetric, nor transitive.
    

Question-3 :-  Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Solution :-
    Let A = {1, 2, 3, 4, 5, 6}.
    A relation R is defined on set A as: R = {(a, b): b = a + 1}
        a = {1, 2, 3, 4, 5}
        b = {2, 3, 4, 5, 6}
  ∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
    We can find (a, a) ∉ R, where a ∈ A.
    For instance, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
  ∴R is not reflexive.
    It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.
  ∴R is not symmetric. 
    Now, (1, 2), (2, 3) ∈ R. But, (1, 3) ∉ R
  ∴R is not transitive.
    Hence, R is neither reflexive, nor symmetric, nor transitive.
    

Question-4 :-  Show that the relation R in defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

Solution :-
    R = {(a, b); a ≤ b}
    Clearly (a, a) ∈ R as a = a. (a is equal to itself)
  ∴R is reflexive.
    Now, (2, 5) ∈ R (as 2 < 5)
    But, (5, 2) ∉ R (as 5 < 2).
  ∴R is not symmetric. 
    Now, let (a, b), (b, c) ∈ R.
    Then, a ≤ b and b ≤ c
 ⇒ a ≤ c
 ⇒ (a, c) ∈ R
  ∴R is transitive.
    Hence,R is reflexive and transitive but not symmetric.
    

Question-5 :-  Check whether the relation R in defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

Solution :-
    R = {(a, b): a ≤ b³}
    It can be observed that a is not equal to square of itself.  
  ∴R is not reflexive.
    Now, (1, 3) ∈ R (as 1 < 3³ = 27)
    But, (3, 1) ∉ R (as 3 < 1³)
  ∴R is not symmetric.
    Now, (3, 2), (2, 1.5) ∈ R
    (as 3 < 2² = 4 and 2 < (1.5)² = 2.25)
    But, 3 > (1.5)² = 2.25
  ∴(3, 1.5) ∉ R
  ∴R is not transitive.
    Hence, R is neither reflexive, nor symmetric, nor transitive.
    

Question-6 :-  Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Solution :-
    Let A = {1, 2, 3}.
    A relation R on A is defined as R = {(1, 2), (2, 1)}.
    It is seen that (1, 1), (2, 2), (3, 3) ∉R.
  ∴R is not reflexive.
    Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric. 
    Now, (1, 2) and (2, 1) ∈ R
    But, (1, 1) ∉ R
  ∴R is not transitive.
    Hence, R is symmetric but neither reflexive nor transitive.
    

Question-7 :-  Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation.

Solution :-
    R = {x, y): x and y have the same number of pages}
    Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.
    Let (x, y) ∈ R ⇒ x and y have the same number of pages.
 ⇒ y and x have the same number of pages.
 ⇒ (y, x) ∈ R
  ∴R is symmetric.
    Now, let (x, y) ∈R and (y, z) ∈ R.
 ⇒ x and y and have the same number of pages and y and z have the same number of pages.
 ⇒ x and z have the same number of pages.
 ⇒ (x, z) ∈ R
  ∴R is transitive.
    Hence, R is an equivalence relation.
    

Question-8 :-  Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Solution :-
    A = {1, 2, 3, 4, 5}
    R = {(a, b) : |a – b|is even}
    It can be observed that a ∈ A and |a – a| = 0 i.e., even no. So,  
  ∴R is reflexive.
    Now, Let (a, b) ∈ R. |a – b|is even.
    and |b – a|is also even. So, (b, a) ∈ R
  ∴R is symmetric.
    Now, (a, b) ∈ R and (b, c) ∈ R. So,
    |a – b|is even and |b – c|is also even.
    (a, b) and (b, c) are even.
    for |a – c|,
    (a - c) = (a - b) + (b - c) sum of two integers is also even. Then,
    |a – c| is even.
  ∴(a, c) ∈ R.
  ∴R is transitive.
    Hence, R is reflexive, symmetric and transitive. 
    So, it is an equivalence relation.

    Now, all elements of the set {1, 2, 3} are related to each other as all the elements of this subset are odd. 
    Thus, the modulus of the difference between any two elements will be even.
    Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
    Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as 
    all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. 
    Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
    

Question-9 :-  Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4}
(ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution :-
    A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(i) R = {(a, b) : |a – b| is a multiple of 4}
    It can be observed that a ∈ A and |a – a| = 0 i.e., is multiple of 4. So,  
  ∴R is reflexive.
    Now, Let (a, b) ∈ R. |a – b|is multiple of 4.
    and |b – a|is also multiple of 4. So, (b, a) ∈ R
  ∴R is symmetric.
    Now, (a, b) ∈ R and (b, c) ∈ R. So,
    |a – b|is multiple of 4 and |b – c|is also multiple of 4.
    (a, b) and (b, c) are multiple of 4.
    for |a – c|,
    (a - c) = (a - b) + (b - c)  is also multiple of 4. Then,
    |a – c| is multiple of 4.
  ∴(a, c) ∈ R.
  ∴R is transitive.
    Hence, R is reflexive, symmetric and transitive. So, 
    It is an equivalence relation.
    Here, The set of elements related to 1 is {1, 5, 9}.
    
(ii) R = {(a, b) : a = b}
    For any element a ∈A, we have (a, a) ∈ R, since a = a.
  ∴R is reflexive. 
    Now, let (a, b) ∈ R.
 ⇒ a = b
 ⇒ b = a
 ⇒ (b, a) ∈ R
  ∴R is symmetric.
    Now, let (a, b) ∈ R and (b, c) ∈ R.
 ⇒ a = b and b = c
 ⇒ a = c
 ⇒ (a, c) ∈ R
  ∴R is transitive.
    Hence, R is an equivalence relation.
    The elements in R that are related to 1 will be those elements from set A which are equal to 1.
    Hence, the set of elements related to 1 is {1}.
    

Question-10 :-  Give an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive

Solution :-
(i) Let A = {1, 2, 3}.
    Define a relation R on A as R = {(1, 2), (2, 3)}.
  ∴R is not reflexive as (1, 1), (2, 2), (3, 3) ∉ R.
    Now, as (1, 2) ∈ R and also (2, 1) ∈ R, 
  ∴R is symmetric. 
    (1, 2), (2, 3) ∈ R, but (1, 3) ∉ R
  ∴R is not transitive.
    Hence, relation R is symmetric but not reflexive or transitive.
    
(ii) Consider a relation R in   defined as: R = {(x, y): x < y}
    For any x ∈ R, we have (x, y) ∉ R since x cannot be strictly less than x itself. In fact, x = x.
  ∴R is not reflexive.
    Now, (x, y) ∈ R (as x < y)
    But, y is not less than x.
  ∴(y, x) ∉ R
  ∴R is not symmetric. 
    Now, let (x, y), (y, z) ∈ R.
 ⇒ x < y and y < z
 ⇒ x < z
 ⇒ (x, z) ∈ R
  ∴R is transitive.
    Hence, relation R is transitive but not reflexive and symmetric.
    
(iii) Let A = {2, 4, 6}.
    Define a relation R on A as:
    A = {(2, 2), (4, 4), (6, 6), (2, 4), (4, 2), (4, 6), (6, 4)}
    Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., {(2, 2), (4, 4), (6, 6)} ∈ R.
    Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R. i.e., {(2, 4), (4, 2), (4, 6), (6, 4)} ∈ R.
    Relation R is not transitive since (2, 4), (4, 6) ∈ R, but (2, 6) ∉ R.
    Hence, relation R is reflexive and symmetric but not transitive.
    
(iv) Define a relation R in    as: R = {a, b): a³ ≥ b³}
    Clearly (a, a) ∈ R as a³ = a³.
  ∴R is reflexive.
    Now, (2, 1) ∈ R (as 2³ ≥ 1³)
    But, (1, 2) ∉ R (as 1³ < 2³)
  ∴R is not symmetric.
    Now, Let (a, b), (b, c) ∈ R.
 ⇒ a³ ≥ b³ and b³ ≥ c³
 ⇒ a³ ≥ c³
 ⇒ (a, c) ∈ R
  ∴R is transitive.
    Hence, relation R is reflexive and transitive but not symmetric.
    
(v) Let A = {−3, −4}.
    Define a relation R on A as:
    R = {(−3, −4), (−4, −3), (−3, −3)}
    Relation R is not reflexive as (−4, −4) ∉ R.
    Relation R is symmetric as (−3, −4) ∈ R and (−4, −3} ∈ R. 
    It is seen that (−3, −4), (−4, −3) ∈ R. Also, (−3, −3) ∈ R.
  ∴The relation R is transitive.
    Hence, relation R is symmetric and transitive but not reflexive.
    

Question-11 :-  Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Solution :-
    R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
    Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the
    distance of the same point P from the origin.
  ∴R is reflexive.
    Now, Let (P, Q) ∈ R.
 ⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.
 ⇒ The distance of point Q from the origin is the same as the distance of point P from the origin. So, 
 ⇒ (Q, P) ∈ R
  ∴R is symmetric.
    Now, Let (P, Q), (Q, S) ∈ R.
 ⇒ The distance of points P and Q from the origin is the same and also, 
    the distance of points Q and S from the origin is the same.
 ⇒ The distance of points P and S from the origin is the same. So,
 ⇒ (P, S) ∈ R
  ∴R is transitive.
    Therefore, R is an equivalence relation.

    The set of all points related to P ≠ (0, 0) will be those points whose distance from the
    origin is the same as the distance of point P from the origin.
    In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P
    is at a distance of k from the origin.
    Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
    

Question-12 :-  Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

Solution :-
    R = {(T1, T2): T1 is similar to T2}
    R is reflexive since every triangle is similar to itself.
    Further, if (T1, T2) ∈ R, then T1 is similar to T2.
 ⇒ T2 is similar to T1. So,
 ⇒ (T2, T1) ∈ R
  ∴R is symmetric.
    Let (T1, T2), (T2, T3) ∈ R.
 ⇒ T1 is similar to T2 and T2 is similar to T3.
 ⇒ T1 is similar to T3. So,
 ⇒ (T1, T3) ∈ R
  ∴R is transitive.
    Thus, R is an equivalence relation.
    Now, we can observe that:
    3/6 = 4/8 = 5/10 = 1/2
  ∴The corresponding sides of triangles T1 and T3 are in the same ratio. 
    Then, triangle T1 is similar to triangle T3.
    Hence, T1 is related to T3.
    

Question-13 :-  Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Solution :-
    R = {(P1, P2): P1 and P2 have same the number of sides}
    R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself.
    Let (P1, P2) ∈ R.
 ⇒ P1 and P2 have the same number of sides.
 ⇒ P2 and P1 have the same number of sides. So,
 ⇒ (P2, P1) ∈ R
  ∴R is symmetric.
    Now, Let (P1, P2), (P2, P3) ∈ R.
 ⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
 ⇒ P1 and P3 have the same number of sides. So,
 ⇒ (P1, P3) ∈ R
  ∴R is transitive.
    Hence, R is an equivalence relation.

    The elements in A related to the rightMangled triangle (T) with sides 3, 4, and 5 are those polygons 
    which have 3 sides (since T is a polygon with 3 sides).
    Hence, the set of all elements in A related to triangle T is the set of all triangles.
    

Question-14 :-  Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Solution :-
    R = {(L1, L2): L1 is parallel to L2}
    R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R. Now,
    Let (L1, L2) ∈ R.
 ⇒ L1 is parallel to L2.
 ⇒ L2 is parallel to L1. So,
 ⇒ (L2, L1) ∈ R
  ∴R is symmetric.
    Now, Let (L1, L2), (L2, L3) ∈ R.
 ⇒ L1 is parallel to L2. Also, L2 is parallel to L3.
 ⇒ L1 is parallel to L3. So,
 ⇒ (L1, L3) ∈ R
  ∴R is transitive.
    Hence, R is an equivalence relation.

    The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.
    Slope of line y = 2x + 4 is m = 2
    It is known that parallel lines have the same slopes.
    The line parallel to the given line is of the form y = 2x + c, where c ∈ R  .
    Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.
    

Question-15 :-  Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.  (A) R is reflexive and symmetric but not transitive.,  (B) R is reflexive and transitive but not symmetric.,  (C) R is symmetric and transitive but not reflexive.,  (D) R is an equivalence relation.

Solution :-
    R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
    It is seen that (a, a) ∈    ! for every a ∈{1, 2, 3, 4}.
  ∴R is reflexive.
    It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
  ∴R is not symmetric.
    Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
  ∴R is transitive.
    Hence, R is reflexive and transitive but not symmetric.
    The correct answer is B.
    

Question-16 :-  Let R be the relation in the set given by R = {(a, b): a = b − 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R,  (B) (3, 8) ∈ R,  (C) (6, 8) ∈ R,  (D) (8, 7) ∈ R.

Solution :-
    R = {(a, b): a = b − 2, b > 6}
    Now, since b > 6, (2, 4) ∉ R
    Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
    And, as 8 ≠ 7 − 2, (8, 7) ∉ R
    Now, consider (6, 8).
    We have 8 > 6 and also, 6 = 8 − 2. ∴(6, 8) ∈ R
    The correct answer is C.
    
CLASSES

Connect with us:

Copyright © 2015-16 by a1classes.

www.000webhost.com