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anonymous
 one year ago
Equivalence relation problem
anonymous
 one year ago
Equivalence relation problem

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444584458933:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for each relation, we need to check : 1) reflexive 2) symmetry 3) transitive

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the reflexive part in i) I am having a confusion

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1to better understand reflexivity, maybe consider a relation that is not reflexive

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1first 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1can you think of a relation that is not reflexive ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1nice, another one : ab is odd

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so do you get why the relation in part i is reflexive ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes now it is clear..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1review quick top 3 properties dw:1444585876121:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1can you guess what the equivalence classes will be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x will be any member of R and y will be fixed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Exactly! for example below is an equivalence class : [(x, 1)] = {(1,1), (2,1), (2.2, 1), (99, 1), ... }

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1below is another equivalence class [(x, 3)] = {(1,3), (2,3), (2.2, 3), (99, 3), ... }

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0[(x,y)]={(a,b)belongs to R^2 such that b=y}

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this is the answer right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes thats the answer for part i

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1look at the relation in part ii, whats ur first guess, can it be an equivalence relation ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it is reflexive and symmetric but transitivity cant say

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1right, just show an example that its not transitive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then we need to consider general points of R^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yes just pick any one simple example

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(1,2)~(1,3) and(1,3)~(2,3) but (1,2) is not ~ to (2,3)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that will do, that proves the relation is not transitive consequently its not equivalence relation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes correct and hence no question of equivalence classes arise

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1good iii looks innocent, but it can be very tricky...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1because 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 ..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1proving that it is an equivalence relation is trivial but figuring out equivalence classes can be tricky...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we see that the y component can be anything

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1hey wait, does it really pass transitivity ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah if ab is an integer and bc is an integer then ac must be an integer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ac=(ab)+(bc)= sum of integers

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh thats clever! okay so it does pass transitivity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes.. and the symmetric and reflexive parts are also satisfied.. so we need to figure out the equivalence realtion

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you mean equivalence `classes`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so our y component can be anything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but x component will be those real numbers whose difference gives integer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1How about this [(x,y)]={(a,b)belongs to R^2 such that a = xfloor(x)+k, \(k \in \mathbb Z\)}

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1floor(x) gives the integer part of x so, x  floor(x) gives the fractional part of x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it gives integer always I thought of another one, [(x,y)]={(a,b)belongs to R^2 such that (xa) belongs to Z}

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1mine is like construction your's is more like a logic statement yeah both works, but yours looks better

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks. so we are done with the questions. I just need to construct the language only..:)
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