Curry
  • Curry
Validation on finding equivalence classes.
Mathematics
katieb
  • katieb
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Curry
  • Curry
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Curry
  • Curry
My answer was n,m) = {(n-(n-1), m+(n-1)), (n-(n-2), m+(n-2)), …}. Am I right?
Loser66
  • Loser66
How can you get it?

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Curry
  • Curry
Well, a equivalence class is all the elements that can go in for a, in aRb.
Curry
  • Curry
But I'm saying that very bluntly. It's more complicated than it sounds? atleast to me.
perl
  • perl
to prove its an eq. class, you have to show the relation is 1. reflexive (a Ra for any a) 2. symmetric (if aRb, then bRa for any a,b) 3. transitive ( if aRb and bRc, then aRc )
perl
  • perl
here a,b,c are elements of NxN
Curry
  • Curry
Well i already showed it's a equivalence class.
Curry
  • Curry
proved the equivalence relation* I just need help finding the equivalence class.
Curry
  • Curry
i figured, if i look at an example, 3R4, then the equivalence classes would include {(1,6),(2,5)}
Curry
  • Curry
right? so, i tried to generalize that. But i'm not sure that's how it goes.
Curry
  • Curry
mhmm, and then?
Loser66
  • Loser66
Then, the equivalence class is all the point on the line y =x
Loser66
  • Loser66
where x, y in N
perl
  • perl
lets see if we can represent the set of members using set notation
Curry
  • Curry
so wait, what does that tell me? that if my solution, with m/n is -1, then it's an equivalence class?
Curry
  • Curry
perl
  • perl
the equivalence classes are the set of all positive ordered pairs, which add up to a number, so for example 2 = {(1,1)} 3= { (1,2) , (2,1) } 4 = { (1,3) , (2,2) (3,1) } , etc.
perl
  • perl
5 = { (1,4) (2,3) (3,2) (4,1) } note that the members of (a,b) must belong to N and N , positive integer
perl
  • perl
unless you are defining N = {0,1,2,3.. } you need to be clear how you are defining N
perl
  • perl
so we can define the equivalence class as follows. [n] = { (a,b) such that a + b = n}
Curry
  • Curry
one second. so ,
Curry
  • Curry
i'd have to define it as for some integer X, [x] = {(a,b) such that a + b = X} ?
perl
  • perl
you don't necessarily need a fancy notation to express the equivalence classes, you just need to describe the equivalence classes sufficiently. If we assume that N = { 1,2,3... } the equivalence classes are : $$\Large {\{ (1,1)\} , ~\{(1,2) , (2,1) \},\\ \{ (1,3) , (2,2), (3,1) \} , ~ \{ (1,4) (2,3) (3,2) (4,1)\} ...}$$
perl
  • perl
you can describe the equivalence classes as pairs of numbers that add up to a given positive integer, starting with 2 , where order counts
perl
  • perl
if you want to define it this way. [x] = {(a,b) a,b are in N , a + b = x }
perl
  • perl
and x is greater than or equal to 2 .
Curry
  • Curry
OOO! that makes a lot of sense! Thanks for spreading all that knowledge! haha
Curry
  • Curry
does (2,3) count as the same thing as (3,2)? or are they 2 different things?
perl
  • perl
if you define N = {0,1,2,3... } then we get the equivalence classes: $$\large { \{(0,0)\}, \\\{ (0,1),(1,0)\} \\ \{ (0,2) , (1,1) , (2,0)\} , \\\{(0,3), (1,2) , (2,1), (3,0)\}, \\ \{ (0,4), (1,3) , (2,2), (3,1), (4,0) \} , \\ \{ (0,5), (1,4) (2,3) (3,2) (4,1), (5,0)\} ... } $$
perl
  • perl
they are two different things, when it comes to 'ordered pairs'
perl
  • perl
also note that the equivalence classes partitions NxN, which is a nice property of equivalence classes
perl
  • perl
partitions NxN into disjoint subsets, which are exhaustive
perl
  • perl
if you graph these equivalence classes, you get lines that move diagonally , and cover the entire quadrant 1, which is NxN
Curry
  • Curry
so we look at each N as separate sets and find every possibility?
perl
  • perl
https://www.desmos.com/calculator/zqufk7le6n
perl
  • perl
each equivalence class is a diagonal line segment of points
perl
  • perl
the positive x axis is N, the positive y axis is N, the points are ordered pairs which represents elements of NxN . Therefore the entire quadrant 1 of discrete points is NxN
perl
  • perl
the 'cartesian product' of NxN
perl
  • perl
you can represent NxN geometrically as the set of 'discrete' points in quadrant 1
perl
  • perl
i think the easiest way to describe the equivalence classes , is that they are ordered pairs that add up to a given positive integer, either starting with 0 if N={0,1,2,3..} or starting with 2 if N={1,2,3,...}
perl
  • perl
here is another well known equivalence relation on NxN, http://mathrefresher.blogspot.com/2006/02/set-of-integers.html
perl
  • perl
using this equivalence relation you can construct negative integers using only positive integers
perl
  • perl
this is a different equivalence relation, not the same as your problem
Curry
  • Curry
i hope I never get another equivalence problem wrong.
Curry
  • Curry
are you cs major by any chance?
perl
  • perl
this question falls within the domain of pure math, i would say. but yeah, it has applications in cs (sort of ~ )
perl
  • perl
also in cs, i think you define N as {0,1,2,3... } though, it kind of depends on how the teacher defines it
perl
  • perl
its too bad there is ambiguity with N. I have seen \( \large \mathbb N^{+} \) to refer to the positive integers
perl
  • perl
http://en.wikipedia.org/wiki/Natural_number#Notation

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