jango_IN_DTOWN
  • jango_IN_DTOWN
Equivalence relation problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jango_IN_DTOWN
  • jango_IN_DTOWN
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jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8
ganeshie8
  • ganeshie8
|dw:1444584458933:dw|

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ganeshie8
  • ganeshie8
for each relation, we need to check : 1) reflexive 2) symmetry 3) transitive
ganeshie8
  • ganeshie8
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
jango_IN_DTOWN
  • jango_IN_DTOWN
the reflexive part in i) I am having a confusion
ganeshie8
  • ganeshie8
to better understand reflexivity, maybe consider a relation that is not reflexive
ganeshie8
  • ganeshie8
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.
ganeshie8
  • ganeshie8
can you think of a relation that is not reflexive ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yes a>b
ganeshie8
  • ganeshie8
nice, another one : a-b is odd
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah
ganeshie8
  • ganeshie8
so do you get why the relation in part i is reflexive ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yes now it is clear..
ganeshie8
  • ganeshie8
review quick top 3 properties |dw:1444585876121:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
checked..
ganeshie8
  • ganeshie8
can you guess what the equivalence classes will be
jango_IN_DTOWN
  • jango_IN_DTOWN
x will be any member of R and y will be fixed
ganeshie8
  • ganeshie8
Exactly! for example below is an equivalence class : [(x, 1)] = {(1,1), (2,1), (2.2, 1), (-99, 1), ... }
ganeshie8
  • ganeshie8
below is another equivalence class [(x, 3)] = {(1,3), (2,3), (2.2, 3), (-99, 3), ... }
jango_IN_DTOWN
  • jango_IN_DTOWN
[(x,y)]={(a,b)belongs to R^2 such that b=y}
ganeshie8
  • ganeshie8
looks nice
jango_IN_DTOWN
  • jango_IN_DTOWN
so this is the answer right?
ganeshie8
  • ganeshie8
Yes thats the answer for part i
ganeshie8
  • ganeshie8
look at the relation in part ii, whats ur first guess, can it be an equivalence relation ?
jango_IN_DTOWN
  • jango_IN_DTOWN
I think it is reflexive and symmetric but transitivity cant say
ganeshie8
  • ganeshie8
right, just show an example that its not transitive
jango_IN_DTOWN
  • jango_IN_DTOWN
then we need to consider general points of R^2
ganeshie8
  • ganeshie8
yes just pick any one simple example
jango_IN_DTOWN
  • jango_IN_DTOWN
(1,2)~(1,3) and(1,3)~(2,3) but (1,2) is not ~ to (2,3)
ganeshie8
  • ganeshie8
that will do, that proves the relation is not transitive consequently its not equivalence relation
jango_IN_DTOWN
  • jango_IN_DTOWN
yes correct and hence no question of equivalence classes arise
ganeshie8
  • ganeshie8
good iii looks innocent, but it can be very tricky...
ganeshie8
  • ganeshie8
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 ..
jango_IN_DTOWN
  • jango_IN_DTOWN
ohhhh
ganeshie8
  • ganeshie8
proving that it is an equivalence relation is trivial but figuring out equivalence classes can be tricky...
jango_IN_DTOWN
  • jango_IN_DTOWN
we see that the y component can be anything
ganeshie8
  • ganeshie8
hey wait, does it really pass transitivity ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah if a-b is an integer and b-c is an integer then a-c must be an integer
jango_IN_DTOWN
  • jango_IN_DTOWN
a-c=(a-b)+(b-c)= sum of integers
ganeshie8
  • ganeshie8
Ahh thats clever! okay so it does pass transitivity
jango_IN_DTOWN
  • jango_IN_DTOWN
yes.. and the symmetric and reflexive parts are also satisfied.. so we need to figure out the equivalence realtion
ganeshie8
  • ganeshie8
you mean equivalence `classes`
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah oops
jango_IN_DTOWN
  • jango_IN_DTOWN
so our y component can be anything
jango_IN_DTOWN
  • jango_IN_DTOWN
but x component will be those real numbers whose difference gives integer
ganeshie8
  • ganeshie8
How about this [(x,y)]={(a,b)belongs to R^2 such that a = x-floor(x)+k, \(k \in \mathbb Z\)}
ganeshie8
  • ganeshie8
floor(x) gives the integer part of x so, x - floor(x) gives the fractional part of x
jango_IN_DTOWN
  • jango_IN_DTOWN
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}
ganeshie8
  • ganeshie8
looks much better!
jango_IN_DTOWN
  • jango_IN_DTOWN
But both will work
ganeshie8
  • ganeshie8
mine is like construction your's is more like a logic statement yeah both works, but yours looks better
jango_IN_DTOWN
  • jango_IN_DTOWN
Thanks. so we are done with the questions. I just need to construct the language only..:)
ganeshie8
  • ganeshie8
np :)

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