Can any of you maths geniuses help me?

soulcrushersoulcrusher
Posted:
in General Discussion edited January 2014
I am stuck in a group theory question, want to give it a try?



Let R be a relation on Z (integers) such that m is a positive integer, x R y if and only if m divides x - y, where x, y are integers.

Prove that this equivalence relation partitions Z into m distinct classes.



[ 11-10-2002: Message edited by: soulcrusher ]</p>

Comments

  • Reply 1 of 6
    alcimedesalcimedes Posts: 5,486member
    sorry man, we don't do homework.



  • Reply 2 of 6
    powerdocpowerdoc Posts: 8,123member
    [quote]Originally posted by alcimedes:

    <strong>sorry man, we don't do homework.



    </strong><hr></blockquote>



    Well played Alcimedes, i have a different reply, but luckily you replied before me (our reputation of Genius is saved ).
  • Reply 3 of 6
    ghost_user_nameghost_user_name Posts: 22,667member
    42 is the answer.
  • Reply 4 of 6
    spaceman_spiffspaceman_spiff Posts: 1,242member
    [quote]Originally posted by Brad:

    <strong>42 is the answer. </strong><hr></blockquote>



    Yes it is. Now where's that damn towel?
  • Reply 5 of 6
    josef k.josef k. Posts: 79member
    Well, the statement of your problem mentions that you've got an equivalence relation, so I assume that's a given. And you mention distinct "classes," so I assume that you've got the definition of an equivalence class to work with as well.



    I would suggest that you forget about the precise statement of the relation R (congruence of integers) and focus instead on equivalence classes in general. Specifically, any equivalence relation on a set A will partition A into mutually disjoint, nonempty subsets. Prove that first, then use the congruence of integers afterwards to prove that there are actually m of these congruence classes.



    You can show that an equivalence relation partitions its set by working with the definition of a partition:



    (Goes to fetch old math textbook)



    Partition:



    Let {A_k}, k in K, be a collection of subsets of the nonempty set A. Then {A_k} is a partition of A iff these conditions apply:



    1. each A_k is nonempty.

    2. A = Union of all A_k over all k in K.

    3. if A_1 intersect A_2 is nonempty, then A_1 = A_2



    For example, the first part is easy to prove: since your equivalence relation is reflexive, you know that xRx for all x in A, so each equivalence class associated with any given element will at least contain itself as a member.



    Use the reflexivity, symmetry and transitivity of R to prove the rest.



    Good luck. (It's been years since I've seen this stuff, btw, so you take my advice at your own risk!)
  • Reply 6 of 6
    agent302agent302 Posts: 974member
    [quote]Originally posted by spaceman_spiff:

    <strong>



    Yes it is. Now where's that damn towel?</strong><hr></blockquote>



    Don't Panic
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