Question

Equivalence relation problem

Answer 1

@ganeshie8

Answer 2

for each relation, we need to check :
1) reflexive
2) symmetry
3) transitive

Answer 3

for i :

reflexive :
\((x_1, y_1)\sim (x_1,y_1)\) because \(y_1=y_1\)

symmetry :
\( (x_1,y_1)\sim(x_2,y_2) \implies (x_2,y_2)\sim (x_1,y_1)\) because \(y_1=y_2 \implies y_2=y_1\)

transitivity :
\( (x_1,y_1)\sim(x_2,y_2) \) and \((x_2,y_2)\sim (x_3,y_3)\) \(\implies (x_1,y_1)\sim (x_3,y_3)\) because \(y_1=y_2\) and \(y_2=y_3\)\(\implies y_1=y_3\)

therefore, this is an equivalence relation

Answer 4

the reflexive part in i) I am having a confusion

Answer 5

to better understand reflexivity,
maybe consider a relation that is not reflexive

Answer 6

first of all, as the name says, "reflexive" refers to the reflection that you see when u look at mirror. you see your own reflection...

we say a relation is reflexive if \(x\sim x\) for all \(x\) in the relation.

Answer 7

can you think of a relation that is not reflexive ?

Answer 8

yes a>b

Answer 9

nice, another one : a-b is odd

Answer 10

yeah

Answer 11

so do you get why the relation in part i is reflexive ?

Answer 12

yes now it is clear..

Answer 13

review quick top 3 properties

Answer 14

checked..

Answer 15

can you guess what the equivalence classes will be

Answer 16

x will be any member of R and y will be fixed

Answer 17

Exactly! for example below is an equivalence class :

[(x, 1)] = {(1,1), (2,1), (2.2, 1), (-99, 1), ... }

Answer 18

below is another equivalence class

[(x, 3)] = {(1,3), (2,3), (2.2, 3), (-99, 3), ... }

Answer 19

[(x,y)]={(a,b)belongs to R^2 such that b=y}

Answer 20

looks nice

Answer 21

so this is the answer right?

Answer 22

Yes thats the answer for part i

Answer 23

look at the relation in part ii,
whats ur first guess, can it be an equivalence relation ?

Answer 24

I think it is reflexive and symmetric but transitivity cant say

Answer 25

right, just show an example that its not transitive

Answer 26

then we need to consider general points of R^2

Answer 27

yes just pick any one simple example

Answer 28

(1,2)~(1,3) and(1,3)~(2,3) but (1,2) is not ~ to (2,3)

Answer 29

that will do, that proves the relation is not transitive
consequently its not equivalence relation

Answer 30

yes correct and hence no question of equivalence classes arise

Answer 31

good
iii looks innocent, but it can be very tricky...

Answer 32

because there are several ways to get an integer by taking difference of two numbers :

3 - 1 = integer
1.4 - 0.4 = integer
0.3 - 0.3 = integer
..

Answer 33

ohhhh

Answer 34

proving that it is an equivalence relation is trivial
but figuring out equivalence classes can be tricky...

Answer 35

we see that the y component can be anything

Answer 36

hey wait, does it really pass transitivity ?

Answer 37

yeah if a-b is an integer and b-c is an integer then a-c must be an integer

Answer 38

a-c=(a-b)+(b-c)= sum of integers

Answer 39

Ahh thats clever! okay so it does pass transitivity

Answer 40

yes.. and the symmetric and reflexive parts are also satisfied.. so we need to figure out the equivalence realtion

Answer 41

you mean equivalence `classes`

Answer 42

yeah oops

Answer 43

so our y component can be anything

Answer 44

but x component will be those real numbers whose difference gives integer

Answer 45

How about this

[(x,y)]={(a,b)belongs to R^2 such that a = x-floor(x)+k, \(k \in \mathbb Z\)}

Answer 46

floor(x) gives the integer part of x

so, x - floor(x) gives the fractional part of x

Answer 47

yeah it gives integer always I thought of another one, [(x,y)]={(a,b)belongs to R^2 such that (x-a) belongs to Z}

Answer 48

looks much better!

Answer 49

But both will work

Answer 50

mine is like construction
your's is more like a logic statement

yeah both works, but yours looks better

Answer 51

Thanks. so we are done with the questions. I just need to construct the language only..:)

Answer 52

np :)