Question

Validation on finding equivalence classes.

Answer 1

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

Answer 2

How can you get it?

Answer 3

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

Answer 4

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

Answer 5

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 )

Answer 6

here a,b,c are elements of NxN

Answer 7

Well i already showed it's a equivalence class.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

mhmm, and then?

Answer 12

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

Answer 13

where x, y in N

Answer 14

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

Answer 15

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

Answer 16

@amistre64 @ganeshie8

Answer 17

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.

Answer 18

5 = { (1,4) (2,3) (3,2) (4,1) }

note that the members of (a,b) must belong to N and N , positive integer

Answer 19

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

Answer 20

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

Answer 21

one second. so ,

Answer 22

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

Answer 23

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 :

$$\Large {\{ (1,1)\} , ~\{(1,2) , (2,1) \},\\ \{ (1,3) , (2,2), (3,1) \} , ~ \{ (1,4) (2,3) (3,2) (4,1)\} ...}$$

Answer 24

you can describe the equivalence classes as pairs of numbers that add up to a given positive integer, starting with 2 , where order counts

Answer 25

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

Answer 26

and x is greater than or equal to 2 .

Answer 27

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

Answer 28

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

Answer 29

if you define N = {0,1,2,3... } then we get the equivalence classes:
$$\large {
\{(0,0)\},
\\\{ (0,1),(1,0)\}
\\ \{ (0,2) , (1,1) , (2,0)\} ,
\\\{(0,3), (1,2) , (2,1), (3,0)\},
\\ \{ (0,4), (1,3) , (2,2), (3,1), (4,0) \} ,
\\ \{ (0,5), (1,4) (2,3) (3,2) (4,1), (5,0)\} ... } $$

Answer 30

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

Answer 31

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

Answer 32

partitions NxN into disjoint subsets, which are exhaustive

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

each equivalence class is a diagonal line segment of points

Answer 37

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

Answer 38

the 'cartesian product' of NxN

Answer 39

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

Answer 40

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,...}

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

i hope I never get another equivalence problem wrong.

Answer 45

are you cs major by any chance?

Answer 46

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

Answer 47

also in cs, i think you define N as {0,1,2,3... }
though, it kind of depends on how the teacher defines it

Answer 48

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

Answer 49

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