1, 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?
Ans:
  Let A =  { 1, 2,3, 4, 5, 6 } 
A relation R is defined on set A  as :
 R =  { (a,b ) : b = a + 1 } 
 ∴ 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.

2, Check whether the relation r in R defined as 
R = { (a, b ) :  a ≤ b³ }  is reflexive, symmetric or transtive ?
Ans: R = { (a, b ) :  a ≤ b³ } 
It is observed that ( ¹/₂, ¹/₂ )  ∉ R  as ¹/₂  >  ( ¹/₂ ) ³ = ¹/₈
  ∴ R is not reflexive.
Now,
(1,2 )  ∈ R ( as 1 < 2³ = 8 )
 But,
 (2,1 ) ∉ R as 2³ > 1 )
 ∴ R is not symmetric.
We have ( 3, ³/₂ ), ( ³/₂, ⁶/₅ ) ∈ R as 
3 < ( ³/₂ ) ³  and  ³/₂ < ( ⁶/₅ ) ³
But  ( 3, ⁶/₅ ) ∉ R  as 3 > ( ⁶/₅ ) ³
∴ R is not  transitive
Hence, R is neither reflexive, nor symmetric, nor transitive. 

3, 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.

Ans: Set A is th set of all books in the library of a college.
 R = { ( x, y ) :  x and y  have 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. 

4, Show that each of the relation R in the set 
  A = { x  ∈ z : 0 ≤ x ≤ 12 } given by 
R = { ( a, b ) :  a = b }  is an equivalence relation.  Find the set of all elements related to 1.
Ans:  R = { ( a, b ) :  a = b }
For  any element  a ∈ A, we  have ( a, a )  ∈ R
   ⇒  a = b
    ⇒  b = a
     ⇒ ( b, a )  ∈ R
  ∴ R is  symmetric. 
Now, let (a, b ) ∈ R ad ( b, c ) ∈ R
   ⇒ a = b and b = c
   ⇒ a = c
    ⇒ ( a, c )  ∈ R
 ∴ R is transitive.
Here, R is an equivalente relation.
The elements in R that are related to 1 will be those elements form set A which are  equal to 1.
Hence, the set of elements related to 1 is { 1 }. 

5, 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 R reflexive and transitive but not symmetric
( c ) R is symmetric and transitive but not reflexive
(d ) R is an equivalente relation 

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

6, Number of binary operations on the set { a, b } are 
 ( A ) 10    ( B ) 16   ( c ) 20   ( D )  8
Ans:  A binary  operation  * on { a, b } is a function form { a, b }  x  { a, b }  →  { a, b }
   ie, *  is a function from { ( a, a ), ( a, b ) , ( b, a )  ( b, b )  →  { a, b }.
Hence, the  total number  of binary operations on the set { a, b }  is  24.  ie, 16.
  The correct answer is B.  

7, Let A = { 1, 2, 3, }.  Then number of equivalence relations containing ( 1, 2 ) is
   ( A) 1   ( B )  2    ( C )  3  ( D )  4
Ans:  It is given that A = { 1, 2, 3 }
The smallest equivalence relaton containing ( 1, 2 ) is given by,
R₁ = { ( 1, 1 ), ( 2, 2 ), ( 3, 3 ), ( 1,2 ) ( 2, 1)
 Now, we are left with only four pairs
   ie, ( 2, 3 ), ( 3, 2 ), ( 1, 3 )  and ( 3, 1 ).
     If we add any one pair [ say ( 2, 3 ) ] to R₁, then for symmetry we must add ( 3, 2 ). Also for transitivity we  are required  to add  ( 1, 3 ) and ( 3, 1 ).
Hence, the only equivalence relation ( bigger than  R₁ ) is the universal relation.
This shows that the total number of equivalente relations containing  ( 1, 2 ) is two.
The correct anwser is B. 

8, Let A = { 1, 2, 3 }.  Then number of relations containing ( 1, 2 ) and ( 1, 3 ) which are reflexive  and symmetric but not transitive is 
( A ) 1   ( B )    2  ( C ) 3   ( D )   4

Ans: The given set is  A = { 1, 2, 3 }.
The smallest relation containing ( 1, 2 ) and ( 1, 3 )  Which is reflexive and symmetric, but not transitive is given by:
   R = { ( 1, 1 ), ( 2, 2) ( 3, 3 ), ( 1, 2 ), ( 1, 3 ),  ( 2, 1 )  ( 3, 1 ) }
This is because relation r is reflexive as ( 1, 1 ) ( 2, 2 ), ( 3,3 ) ∈ R.
Relation R is symmetric since ( 1, 2 ), ( 2, 1 ) ∈ R and    ( 1, 3 ) ( 3, 1 ) ∈ R.
But relation R is not  transitive as ( 3, 1 ) ( 1, 2 ) ∈ R   but  ( 3, 2 )  ∉ R,
Now, If we add t any two pairs ( 3, 2 ) and ( 2, 3 )  ( or both )  to relation R, the relation R will become transitive.
Hence, the total number of desired relations is one.  
The correct answer is  A.

9, Let S = { a, b , c } and T = { 1, 2, 3 }.  Find F -1 of the following functins  F from  S to -1 if it exists.
   ( i ) F = { (a, 3 ), ( b, 2 ), (c, 1 ) } 
   (ii) F = { ) a, 2 ), ( b, 1 ), ( c, 1 )}

Ans:  S = { a, b, c}, T = { 1, 2, 3 }
 ( i )  F : S →  T  is defined as :
F = { ( a, 3 ), ( b, 2 ), (c, 1 ) } 
      ⇒ F ( a )  = 3, F ( b ) = 2,  F ( C ) = 1
   ∴  F^-1  : T  → S is defined as :
  F ^-1  = { (3, a ), (2, b ),  ( 1, C )}.
(ii) F : S → T is defined as :
  F = { (a, 2 ), ( b, 1 ), ( c, 1 )}
Since F ( b ) = F ( c ) = 1, F is not  one - one 
Hence, F is not invertible
 ie, F ^-1 does not exist. 

10, Find the numbr of all Onto function from the Set { 1, 2, 3, - - - -  n ) to itself.
Ans:
Onto functions from the set  { 1, 2, 3, -  - -  , n } to itself is simply a permutation on n symbols 1, 2,  - - - - , n
  Thus. The total number  of onto maps from { 1, 2, - - - - ,n } of permutations on  n symbols  1, 2,  -- - - - n, which is n.