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Equivalence Classes

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Let A be the set, A = {0, 1, 2, 3}.

    Consider the relation R on A give by xRy iff there exists [itex] k \in \mathbb{Z} [/itex] such that x - y = 3k.

    Describe the equivalence classes of R.

    2. Relevant equations



    3. The attempt at a solution

    [itex] E_x = \{y \in A: y - x = 3k, k \in \mathbb{Z} \} [/itex]

    For y = 0, k = -x/3

    y = 1, k = (1-x)/3

    y = 2, k = (2 - x)/3

    y = 3, k = (3 - x)/4

    Therefore there are four equivalence classes?

    I have tried and tried to understand equivalence classes to no avail. I understand the analogies, but then when it transfers to numbers it's just completely lost on me. Even if what I've written above is correct, I just can't wrap my mind around what it means. I'm at one of those points where math makes you feel like the most worthless human to ever live.
     
  2. jcsd
  3. May 2, 2013 #2

    Zondrina

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    "Being a mathematician is a bit like being a manic depressive. You spend your life alternating between giddy elation and black despair" - Steven Krantz

    Not every day is a good day. There will be some days where things may not seem to make sense or it may be hard to formulate reasoning, but tomorrow is another day.

    As for your problem. What is it you don't understand? I can also tell you re-wrote the rule so you could get everything in terms of x's.
     
  4. May 2, 2013 #3
    I guess I am lost in multiple concepts.

    R is a relation on A given by xRy. Given that A only contains 0, 1, 2, and 3, does that mean that x and y can only take on these values? The relation x - y = 3k defines how x is related to y?

    So say y = 0. Then the relation is saying that x is related to 0 when k = -x/3? or should it be written in terms of k: x = -3k?

    I guess I'm really confused right here. x can only take on the value of 0, 1, 2, or 3. So that means k must be 0, -1/3, -2/3, or -1 to satisfy the relation? I don't really understand what this is saying. I understand that equivalence classes delineate different classes of the relation (maybe?), but I'm just not seeing this delineation. All I see is a giant mess of meaningless relationships.

    If y = 0, then k = -x/3. So k = 0, -1/3, -2/3, -1. (for x = 0, 1, 2, 3; respectively)
    If y = 1, then k = (1 - x)/3. So k = 1/3, 0, -1/3, -2/3. (for x = 0, 1, 2, 3; respectively)
    If y = 2, then k = (2 - x)/3. So k = 2/3, 1/3, 0, -1/3. (for x = 0, 1, 2, 3; respectively)
    If y = 3, then k = (3 - x)/3. So k = 1, 2/3, 1/3, 0. (for x = 0, 1, 2, 3; respectively)

    The Krantz quote is great, by the way.
     
  5. May 2, 2013 #4

    Dick

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    Just think really simple. k has to be an integer. Which elements of A are related to 0? You are writing way too much.
     
  6. May 2, 2013 #5
    So 0R0, 0R3; 1R1, and 2R2?
     
  7. May 2, 2013 #6

    Zondrina

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    Yeah it's a quote that's stuck with me and I never forget it because of what it allowed me to realize.

    Wiki has a really great definition of equivalence classes that's easy to understand i'll post it here if it helps you any :

    I highlighted the important parts. This may or may not help you, but it's here for reference.

    Also note that the equivalence class of any element of your set is defined as :

    $$[a] = \{ x \in X \space | \space aRx \}$$

    So lets do one for 0 from your set A.

    ##[0] = \{ a \in A \space | \space 0Ra \}##

    Does this help?
     
  8. May 2, 2013 #7

    Dick

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    Yes. So what are the equivalence classes? There are three of them.
     
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