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Curry

  • one year ago

Validation on finding equivalence classes.

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

  3. Loser66
    • one year ago
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    How can you get it?

  4. Curry
    • one year ago
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    Well, a equivalence class is all the elements that can go in for a, in aRb.

  5. Curry
    • one year ago
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    But I'm saying that very bluntly. It's more complicated than it sounds? atleast to me.

  6. perl
    • one year ago
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    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 )

  7. perl
    • one year ago
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    here a,b,c are elements of NxN

  8. Curry
    • one year ago
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    Well i already showed it's a equivalence class.

  9. Curry
    • one year ago
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    proved the equivalence relation* I just need help finding the equivalence class.

  10. Curry
    • one year ago
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    i figured, if i look at an example, 3R4, then the equivalence classes would include {(1,6),(2,5)}

  11. Curry
    • one year ago
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    right? so, i tried to generalize that. But i'm not sure that's how it goes.

  12. Curry
    • one year ago
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    mhmm, and then?

  13. Loser66
    • one year ago
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    Then, the equivalence class is all the point on the line y =x

  14. Loser66
    • one year ago
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    where x, y in N

  15. perl
    • one year ago
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    lets see if we can represent the set of members using set notation

  16. Curry
    • one year ago
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    so wait, what does that tell me? that if my solution, with m/n is -1, then it's an equivalence class?

  17. Curry
    • one year ago
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    @amistre64 @ganeshie8

  18. perl
    • one year ago
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    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.

  19. perl
    • one year ago
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    5 = { (1,4) (2,3) (3,2) (4,1) } note that the members of (a,b) must belong to N and N , positive integer

  20. perl
    • one year ago
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    unless you are defining N = {0,1,2,3.. } you need to be clear how you are defining N

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

  22. Curry
    • one year ago
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    one second. so ,

  23. Curry
    • one year ago
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    i'd have to define it as for some integer X, [x] = {(a,b) such that a + b = X} ?

  24. perl
    • one year ago
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    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 : $$\Large {\{ (1,1)\} , ~\{(1,2) , (2,1) \},\\ \{ (1,3) , (2,2), (3,1) \} , ~ \{ (1,4) (2,3) (3,2) (4,1)\} ...}$$

  25. perl
    • one year ago
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    you can describe the equivalence classes as pairs of numbers that add up to a given positive integer, starting with 2 , where order counts

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

  27. perl
    • one year ago
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    and x is greater than or equal to 2 .

  28. Curry
    • one year ago
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    OOO! that makes a lot of sense! Thanks for spreading all that knowledge! haha

  29. Curry
    • one year ago
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    does (2,3) count as the same thing as (3,2)? or are they 2 different things?

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

  31. perl
    • one year ago
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    they are two different things, when it comes to 'ordered pairs'

  32. perl
    • one year ago
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    also note that the equivalence classes partitions NxN, which is a nice property of equivalence classes

  33. perl
    • one year ago
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    partitions NxN into disjoint subsets, which are exhaustive

  34. perl
    • one year ago
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    if you graph these equivalence classes, you get lines that move diagonally , and cover the entire quadrant 1, which is NxN

  35. Curry
    • one year ago
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    so we look at each N as separate sets and find every possibility?

  36. perl
    • one year ago
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    https://www.desmos.com/calculator/zqufk7le6n

  37. perl
    • one year ago
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    each equivalence class is a diagonal line segment of points

  38. perl
    • one year ago
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    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

  39. perl
    • one year ago
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    the 'cartesian product' of NxN

  40. perl
    • one year ago
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    you can represent NxN geometrically as the set of 'discrete' points in quadrant 1

  41. perl
    • one year ago
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    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,...}

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

  43. perl
    • one year ago
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    using this equivalence relation you can construct negative integers using only positive integers

  44. perl
    • one year ago
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    this is a different equivalence relation, not the same as your problem

  45. Curry
    • one year ago
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    i hope I never get another equivalence problem wrong.

  46. Curry
    • one year ago
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    are you cs major by any chance?

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

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

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

  50. perl
    • one year ago
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    http://en.wikipedia.org/wiki/Natural_number#Notation

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