Validation on finding equivalence classes.

- Curry

Validation on finding equivalence classes.

- katieb

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- Curry

##### 1 Attachment

- Curry

My answer was n,m) = {(n-(n-1), m+(n-1)), (n-(n-2), m+(n-2)), …}. Am I right?

- Loser66

How can you get it?

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## More answers

- Curry

Well, a equivalence class is all the elements that can go in for a, in aRb.

- Curry

But I'm saying that very bluntly. It's more complicated than it sounds? atleast to me.

- 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

here a,b,c are elements of NxN

- Curry

Well i already showed it's a equivalence class.

- Curry

proved the equivalence relation* I just need help finding the equivalence class.

- Curry

i figured, if i look at an example, 3R4, then the equivalence classes would include {(1,6),(2,5)}

- Curry

right? so, i tried to generalize that. But i'm not sure that's how it goes.

- Curry

mhmm, and then?

- Loser66

Then, the equivalence class is all the point on the line y =x

- Loser66

where x, y in N

- perl

lets see if we can represent the set of members using set notation

- Curry

so wait, what does that tell me? that if my solution, with m/n is -1, then it's an equivalence class?

- Curry

- 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

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

unless you are defining N = {0,1,2,3.. }
you need to be clear how you are defining N

- perl

so we can define the equivalence class as follows.
[n] = { (a,b) such that a + b = n}

- Curry

one second. so ,

- Curry

i'd have to define it as for some integer X, [x] = {(a,b) such that a + b = X} ?

- 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

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

if you want to define it this way.
[x] = {(a,b) a,b are in N , a + b = x }

- perl

and x is greater than or equal to 2 .

- Curry

OOO! that makes a lot of sense! Thanks for spreading all that knowledge! haha

- Curry

does (2,3) count as the same thing as (3,2)? or are they 2 different things?

- 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

they are two different things, when it comes to 'ordered pairs'

- perl

also note that the equivalence classes partitions NxN, which is a nice property of equivalence classes

- perl

partitions NxN into disjoint subsets, which are exhaustive

- perl

if you graph these equivalence classes, you get lines that move diagonally , and cover the entire quadrant 1, which is NxN

- Curry

so we look at each N as separate sets and find every possibility?

- perl

https://www.desmos.com/calculator/zqufk7le6n

- perl

each equivalence class is a diagonal line segment of points

- 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

the 'cartesian product' of NxN

- perl

you can represent NxN geometrically as the set of 'discrete' points in quadrant 1

- 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

here is another well known equivalence relation on NxN,
http://mathrefresher.blogspot.com/2006/02/set-of-integers.html

- perl

using this equivalence relation you can construct negative integers using only positive integers

- perl

this is a different equivalence relation, not the same as your problem

- Curry

i hope I never get another equivalence problem wrong.

- Curry

are you cs major by any chance?

- perl

this question falls within the domain of pure math, i would say. but yeah, it has applications in cs (sort of ~ )

- 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

its too bad there is ambiguity with N.
I have seen \( \large \mathbb N^{+} \) to refer to the positive integers

- perl

http://en.wikipedia.org/wiki/Natural_number#Notation

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