relations and functions

[kauser]

I need answers for miscellaneous Q. 15, 16. 17, 18 ,

and in exercise 1.4 Q.10, is it applicable for all questions of Q.9

are all the examples equally likely to be asked, or should more emphasis be given to the exercise questions.....BTW the sample papers have all text questions...

[pkciit]

no it is not applicable to all of que.9
some of them have identity

[pkciit]

1,2,3,4 does not have any identity
5 = it has a identity 4
6 = has a identity [+-]1

[sachinsethi]

hey hi. here are the answers..

ex 1.4 10. i. identity cannot exist cause for any element.. a - e is not equal to e - a
ii. always.. a^2 +b^2 will be positive and gr8er than a. if 0 is chosen.. the also.. identity cannot exist for all as a^2 will be gr8er than zero for all but a =1. hence.. identity does not exist.
iii. a + ae is not equal to e+ea for all elements.. hence no identity
iv clearly.. no identity in this case.. cauz always (xcept for a=1) a^2 is greater than a. so no identiy.
v. this is where the ques goes wrong.. identity exists.. and is equal to 4. this is easy to verify.
vi . no identity actually.. cauz even if u choose +1 or -1. a(1^2) is not equal to 1(a^2) for all elements.. similarly for -1.

[sachinsethi]

misc.
15. see for every element in A... f (a) = g(a)
like f(-1) = g(-1) = 0
f(2) = g(2) = 2

thus... the functions hav to be equal.. u hav to show the equality for each element... and thts it.

16. only for { (1,3) , (1,2) (1,1) , (2,2), (3,3) , (3,1), (2,1) } thus answer is A if u add (2,3) u will hav to add (3,2) and tht will make the function transitive, it is not transitive here... as (2,1) and (1,3) are present but not (2,3)

17. equivalence relations = ( (1,1) , (2,2) , (3,3) , (1,2), (2,1) } and then the universal set wid all elements..

18. see.. in above context wrt ques..
let us suppose element is x = 0.4
then fog = f(g(x))
f(g(0.4)) = f([0.4])= f(0) = 0

and gof = g(f(x)) = g(f(0.4)) = g(1)= [1] = 1

hance... they don't coincide !!

[kauser]

Sachin thanx alot

I have a doubt in Q. 16 17

in 17 to show its an equi relation for trans we take (1,2) (2,1) (1,1)

but in Q16 we dont want it to be trans but still we have the same kindo elements as in Q17


and is it a must that all terms must satisfy the condition like if we have to show its reflexive adn we have (1,1) adn not (2,2)

where 1 and 2 belong to R the set , will it still be reflexive ???? I hope you understand what i am saying.

in 1.4 Q.8 why is there no identity for it?


thanx again

[kauser]

Q. 1. Let A = {1,2,3,4} find the number of relations on A containing (1, 2) and (3, 2) which is reflexive transitive but not symmetric giving sufficient reasons.

Q. 2. Let R be the set of real numbers on ‘R’ defined by R = {(a,b) / | a-b | < 5} is not transitive . Prove the above statement by giving two examples.

[ANS (3,6) (6,9) (3,9)
(2,4) (4,8) (2,8) ]

Q. 7. Let f: N --> Y be a function defined by f(x) = x2 + 1 Show that f is one – one and replace Y by a set so that f is invertible. Also find its inverse function.
Q. 8. Let f : R -> R g: R -> R defined by
a. f(x)=x2 + 8 g(x)=3x3 + 1
b. f(x)=x2 + 2x – 3 g(x) = 3x – 4 Show that fog and gof exists and hence find them.
Q. 2. The minimum number of elements that must be added to the relation R = { (1,2), (2,3) } on the set of natural numbers so that it is an equivalence is
a. 4
b. 7 IS THIS THE ANS?
c. 6
d. 5
Q. 3. Let ‘R’ be a reflexive relation on a finite set ‘A’ having ‘n’ elements and let there be ‘m’ ordered pairs in ‘R’. Then
a.
b.
c.
d. none of these.
Q. 4. Let ‘X’ be a family of sets and ‘R’ be a relation defined by ‘A is disjoint from B’ .Then R is
a. Reflexive
b. Symmetric
c. Anti symmetric
d. Transitive
Q. 5. The number of surjections from A = {1, 2, 3, ….. n} on to B = {a, b} is
a. nP2
b. 2n - 2
c. 2n - 1
d. none


Q. 9. Let f(x) = [x] and g(x) = x - [x] then which of the following function is the zero function
a. (f+g) (x)
b. (fg) (x)
c. (f-g) (x)
d. fog (x)

[sachinsethi]

see for this case...
{(1,3) , (1,2) (1,1) , (2,2), (3,3) , (3,1), (2,1) }
is is true.. tht (1,2), (2,1) and (1,1) is a trans prop..
but the relation will be transitive onli if it satisfies for ALL such elements so tht if (a,b) and (b,c) belongs to R , (a,c) wil have to belong to R also..
thus.. to make is transitive.. (2,3) has to be present... because.,. (3,1) and (1,2) are present

[kauser]

thanx