Question

Consider the division of set A = {1,2,3,4,5,6,7,8} by subsets {1,6},{2,7},{3,8},{4},{5}. Define a relation in A by R = {(a,b) : a and b lie in the same subset of the division of A}. Show that R is an equivalence relation.

Answer 1

you have 3 things to show

Answer 2

what are those sir

Answer 3

reflexive,symmetric, transitive

Answer 4

yes that will make R equivalence right ?

Answer 5

if all 3 hold...yes

Answer 6

but what about define a relation in A by R =............... and so

Answer 7

?

Answer 8

they give you the definition....use that definition to show the 3 things

Answer 9

but what about define a relation in A by R =............... and so

Answer 10

i mean to say that the part of the question " Define a relation in A by R = {(a,b) : a and b lie in the same subset of the division of A}." is a question or not ?

Answer 11

but what about define a relation in A by R =............... and so

Answer 12

sorry couldn't understand this" and when u hit 4 u start to go above the parameters given correct"

Answer 13

just show for any \[a\in A\]

\[(a,a)\in R\]

Answer 14

but what about define a relation in A by R =............... and so

Answer 15

how ?

Answer 16

then for any \[(a,b)\in R\]

you have \[(b,a)\in R\]

Answer 17

but what about define a relation in A by R =............... and so

Answer 18

but how will i show them sir ?

Answer 19

you are over thinking this

Answer 20

but what about define a relation in A by R =............... and so

Answer 21

??

Answer 22

this is the definition of R

R = {(a,b) : a and b lie in the same subset of the division of A}

Answer 23

pick any number from 1 to 8

Answer 24

but what about define a relation in A by R =............... and so

Answer 25

yes it is given in the question. But what to do then ?

Answer 26

call it a

if a is in a set is a also in that set :)

Answer 27

but what about define a relation in A by R =............... and so

Answer 28

ok lets take (1,6)

Answer 29

help me also see qus

Answer 30

@zarkon r u posing as zarkon from voltron ?

Answer 31

I am Zarkon from Voltron

Answer 32

sorry

Answer 33

but what about define a relation in A by R =............... and so

Answer 34

Sir please help me.

Kindly post the answer in a single post

Answer 35

help me

Answer 36

but what about define a relation in A by R =............... and so

Answer 37

What is the prob.?

Answer 38

kush help me

Answer 39

but what about define a relation in A by R =............... and so

Answer 40

k i will help u .

Answer 41

and all of u

Answer 42

I don't understand why you keep posting the same thing...

after the word 'by' it gives the definition of R...that is what it is. you are not defining anything.

Answer 43

yes u can help me

Answer 44

but what about define a relation in A by R =............... and so

Answer 45

sir it is not my mistake it is the mistake of my internet

Answer 46

but what about define a relation in A by R =............... and so

Answer 47

sorry !! Please don't mind those sentences sir u continue in the solution

Answer 48

skip reflexive...try symmetric


...

Answer 49

of example \[(1.6)\in R\]

correct?

Answer 50

but what about define a relation in A by R =............... and so

Answer 51

yes

Answer 52

is \[(6,1)\in R\]

Answer 53

but what about define a relation in A by R =............... and so

Answer 54

why sir ?

Answer 55

in other words...are 6 and 1 in the same subset?

Answer 56

but what about define a relation in A by R =............... and so

Answer 57

yes

Answer 58

but what about define a relation in A by R =............... and so

Answer 59

i got that sir . So R is symmetric. Then ?

Answer 60

the symmetry holds for (1,6) and (6,1)

Answer 61

but what about define a relation in A by R =............... and so

Answer 62

yes sir

Answer 63

you would really need to show it is true for all pairs

Answer 64

but what about define a relation in A by R =............... and so

Answer 65

U mean to say for (1,6) and (6,1) then (2,7) and (7,2) right ?

Answer 66

(n,n+5) and (n+5,n) are symmetric for n=1,2,3

Answer 67

yes...that would work too

Answer 68

but what about define a relation in A by R =............... and so

Answer 69

Okay

Answer 70

but what about define a relation in A by R =............... and so

Answer 71

but what about (4) , (5) ?

Answer 72

they don't have pairs so they will be just covered under reflexivity

Answer 73

but what about define a relation in A by R =............... and so

Answer 74

i.e. \[(4,4)\in R \]

Answer 75

yes

Answer 76

\[(n,n)\in R\]

for \[n=1,2,\ldots,7,8\]

Answer 77

but what about define a relation in A by R =............... and so

Answer 78

okay now transitive ?

Answer 79

normally you would need something like this

if (a,b) in R and (b,c) in R then (a,c) is in R

Answer 80

but our sets are so small we only have limited options

Answer 81

for example...

if (1,6) in R and (6,1) in R is (1,1) in R?

Answer 82

but what about define a relation in A by R =............... and so

Answer 83

yes

Answer 84

but what about define a relation in A by R =............... and so

Answer 85

Okay I will try it.

Answer 86

yes...since we already have that by reflexivity :)

Answer 87

but what about define a relation in A by R =............... and so

Answer 88

yes

Answer 89

If f(x) = (5x + 3)/(4x-5), x is not equal to 5/4, show that fof is an identity function.

Answer 90

Sir next question