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Question 6. Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

test

Q. 19 Number of binary operations on the set are
(A) 10
(B) 16
(C) 20
(D ) 8

Binary operations on the set are is a function from   i.e. * is a function from    Hence, the total number of binary operations on set   is  Hence, option B is correct.

Q. 18 Let be the Signum Function defined as

and  be the Greatest Integer Function given by , where  is greatest integer less than or equal to . Then, does and coincide in ?

is defined as    is defined as  Let     Then we have ,  if x=1   and                                                            Hence,for  ,     and . Hence , gof and fog do not coincide with .

Q. 17 Let . Then number of equivalence relations containing  is
(A) 1
(B) 2
(C) 3
(D) 4

The number of equivalence relations containing  is given by  We are left with four  pairs  ,, .   , so relation R is reflexive.     and     , so relation R is not symmetric.    , so realation R is not transitive. Hence , equivalence relation is bigger than R  is universal relation. Thus the total number of equivalence relations cotaining   is two. Thus, option B is correct.

Q.16 Let . Then number of relations containing and which are
reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

The smallest  relations containing and which are reflexive and symmetric but not transitive is given by    , so relation R is reflexive.     and     , so relation R is symmetric.   but   , so realation R is not transitive. Now, if we add any two pairs    and  to relation R, then relation R  will become transitive. Hence, the total number of the desired relation is one. Thus, option A is...

Q. 15 Let ,  and be functions defined  by  and . Are  and  equal?

such that , are called equal functions).

Given : ,       are defined  by  and . It can be observed that                                 Hence, f and g are equal functions.

Q. 14  Define a binary operation ∗ on the set as

Show that zero is the identity for this operation and each element of the set
is invertible with being the inverse of .

X = as    An element     is identity element for operation *,  if  For  ,                                             Hence, 0 is identity element of operation *. An element    is invertible if there exists ,  such that         i.e.     means       or     But since we have  X =  and . Then .    is inverse of a  for . Hence, inverse of element  ,   is  6-a   i.e. ,

Q. 13   Given a non-empty set X, let be defined as
Show that the empty set  is the
identity for the operation ∗ and all the elements A of P(X) are invertible with
. (Hint : and ).

Let   be defined as          Let  . Then                                                                 for all    Thus,  is identity element for operation *. An element     will be invertible if there exists   , such that .     (here  is identity element)       Hence, all elements A  of P(X) are invertible with

Q 13 ) Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

It is given that   be defined as Now, let  . Then,  And Therefore, Therefore, we can say that    is the identity element for the given operation *. Now, an element A  P(X) will be invertible if there exists B  P(X) such that Now, We can see that   such that  Therefore, by this we can say that all the element A of P(X) are invertible with

Q. 12  Consider the binary operations and defined as
and . Show that ∗ is commutative but not
associative, is associative but not commutative. Further, show that ,
. [If it is so, we say that the operation ∗ distributes

Given     and    is defined as                and    For  , we have                                                                                                                          the operation is commutative.                              where                     the operation is not associative Let   . Then we have :                                                               ...

Q. 11 Let  and . Find of the following functions F from S
to T, if it exists.

(ii)

is defined as      , F is not one-one. So inverse of F does not exists. Hence, F is not invertible i.e.  does not exists.

Q. 11 Let and . Find  of the following functions F from S
to T, if it exists.

(i)

is defined as      is given by

Q. 10 Find the number of all onto functions from the set to itself.

The number of all onto functions from the set to itself is  permutations on n symbols  1,2,3,4,5...............n.  Hence, permutations  on n symbols  1,2,3,4,5...............n = n Thus, total number of all onto maps from the set to itself is same as  permutations on n symbols  1,2,3,4,5...............n which is  n.

Q. 9 Given a non-empty set X, consider the binary operation
given by , where P(X) is the power set of X.
Show that X is the identity element for this operation and X is the only invertible
element in P(X) with respect to the operation ∗.

Given      is defined as  . As we know that                  Hence, X is the identity element of binary operation *. Now, an element  is invertible if there exists a  , such that              (X is identity element)   i.e.              This is possible only if  . Hence, X is only invertible element  in  with respect to operation *                                                                 ...

Q.8 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if . Is R an equivalence relation

Given a non empty set X, consider P(X) which is the set of all subsets of X. Since, every set is subset of itself , ARA  for all    R is reflexive. Let  This is not same as  If        and     then we cannot say that B is related to A.  R is not symmetric. If  this implies     R is transitive. Thus, R is not an equivalence relation because it is not symmetric.

Q. 7 Give examples of two functions and  such that is onto
but  is not onto.

(Hint : Consider  and

and               and         Onto :            Consider element in codomain N . It is clear that this element is not an image of any of element  in domain N .     f is not onto.      Now, it is clear that   , there exists       such that  . Hence,  is onto.

Q. 6 Give examples of two functions and  such that is
injective but g is not injective.

(Hint : Consider and ).

One - one: Since     As we can see    but   so    is not one-one. Thus , g(x) is not injective. Let                             Since,     so x and y are both positive.          Hence, gof is injective.

Q.5 Show that the function  given by is injective.

One-one:    Let                   We need to prove .So, Let     then there cubes will not be equal i.e. .  It will contradict given condition of cubes being equal.  Hence,   and it is one -one which means it is injective.

Q. 4 Show that the function defined by      is one one and onto function.

The function    defined by    ,  One- one:          Let                  ,                      It is observed that if x is positive and y is negative.           Since x is positive and y is negative.     but 2xy is negative. Thus, the case of x is positive and y is negative is removed. Same happens in the case  of y is positive and x is negative so this case is also removed. When x and y...

Q. 3 If f : R → R is defined by f(x) = x 2– 3x + 2, find f (f (x)).

This can be solved as following f : R → R
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