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Curry
 one year ago
Validation on finding equivalence classes.
Curry
 one year ago
Validation on finding equivalence classes.

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Curry
 one year ago
Best ResponseYou've already chosen the best response.1My answer was n,m) = {(n(n1), m+(n1)), (n(n2), m+(n2)), …}. Am I right?

Curry
 one year ago
Best ResponseYou've already chosen the best response.1Well, a equivalence class is all the elements that can go in for a, in aRb.

Curry
 one year ago
Best ResponseYou've already chosen the best response.1But I'm saying that very bluntly. It's more complicated than it sounds? atleast to me.

perl
 one year ago
Best ResponseYou've already chosen the best response.2to 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
 one year ago
Best ResponseYou've already chosen the best response.2here a,b,c are elements of NxN

Curry
 one year ago
Best ResponseYou've already chosen the best response.1Well i already showed it's a equivalence class.

Curry
 one year ago
Best ResponseYou've already chosen the best response.1proved the equivalence relation* I just need help finding the equivalence class.

Curry
 one year ago
Best ResponseYou've already chosen the best response.1i figured, if i look at an example, 3R4, then the equivalence classes would include {(1,6),(2,5)}

Curry
 one year ago
Best ResponseYou've already chosen the best response.1right? so, i tried to generalize that. But i'm not sure that's how it goes.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Then, the equivalence class is all the point on the line y =x

perl
 one year ago
Best ResponseYou've already chosen the best response.2lets see if we can represent the set of members using set notation

Curry
 one year ago
Best ResponseYou've already chosen the best response.1so wait, what does that tell me? that if my solution, with m/n is 1, then it's an equivalence class?

perl
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.25 = { (1,4) (2,3) (3,2) (4,1) } note that the members of (a,b) must belong to N and N , positive integer

perl
 one year ago
Best ResponseYou've already chosen the best response.2unless you are defining N = {0,1,2,3.. } you need to be clear how you are defining N

perl
 one year ago
Best ResponseYou've already chosen the best response.2so we can define the equivalence class as follows. [n] = { (a,b) such that a + b = n}

Curry
 one year ago
Best ResponseYou've already chosen the best response.1i'd have to define it as for some integer X, [x] = {(a,b) such that a + b = X} ?

perl
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2you can describe the equivalence classes as pairs of numbers that add up to a given positive integer, starting with 2 , where order counts

perl
 one year ago
Best ResponseYou've already chosen the best response.2if you want to define it this way. [x] = {(a,b) a,b are in N , a + b = x }

perl
 one year ago
Best ResponseYou've already chosen the best response.2and x is greater than or equal to 2 .

Curry
 one year ago
Best ResponseYou've already chosen the best response.1OOO! that makes a lot of sense! Thanks for spreading all that knowledge! haha

Curry
 one year ago
Best ResponseYou've already chosen the best response.1does (2,3) count as the same thing as (3,2)? or are they 2 different things?

perl
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.2they are two different things, when it comes to 'ordered pairs'

perl
 one year ago
Best ResponseYou've already chosen the best response.2also note that the equivalence classes partitions NxN, which is a nice property of equivalence classes

perl
 one year ago
Best ResponseYou've already chosen the best response.2partitions NxN into disjoint subsets, which are exhaustive

perl
 one year ago
Best ResponseYou've already chosen the best response.2if you graph these equivalence classes, you get lines that move diagonally , and cover the entire quadrant 1, which is NxN

Curry
 one year ago
Best ResponseYou've already chosen the best response.1so we look at each N as separate sets and find every possibility?

perl
 one year ago
Best ResponseYou've already chosen the best response.2each equivalence class is a diagonal line segment of points

perl
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.2the 'cartesian product' of NxN

perl
 one year ago
Best ResponseYou've already chosen the best response.2you can represent NxN geometrically as the set of 'discrete' points in quadrant 1

perl
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.2here is another well known equivalence relation on NxN, http://mathrefresher.blogspot.com/2006/02/setofintegers.html

perl
 one year ago
Best ResponseYou've already chosen the best response.2using this equivalence relation you can construct negative integers using only positive integers

perl
 one year ago
Best ResponseYou've already chosen the best response.2this is a different equivalence relation, not the same as your problem

Curry
 one year ago
Best ResponseYou've already chosen the best response.1i hope I never get another equivalence problem wrong.

Curry
 one year ago
Best ResponseYou've already chosen the best response.1are you cs major by any chance?

perl
 one year ago
Best ResponseYou've already chosen the best response.2this question falls within the domain of pure math, i would say. but yeah, it has applications in cs (sort of ~ )

perl
 one year ago
Best ResponseYou've already chosen the best response.2also in cs, i think you define N as {0,1,2,3... } though, it kind of depends on how the teacher defines it

perl
 one year ago
Best ResponseYou've already chosen the best response.2its too bad there is ambiguity with N. I have seen \( \large \mathbb N^{+} \) to refer to the positive integers
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