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Relations and partitions

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data
    (proof) Determine whether or not (x,y)~(w,z) if and only if y=w is an equivalence relation. If it is, then describe the associated partition.


    2. Relevant equations



    3. The attempt at a solution

    Let x be an element of the reals. It is known that a relation on a set X is said to be reflexive if x~x for all x is an element of X. Since y and w are equal and x is an element of the reals and X is an element of the reals, this relation is reflexive.

    It is known that a relaion on a set X is symmetric if for all x, y is an element of X, whenever x~y, then y~x. Since y=w, which is the same thing as saying, x~y and y~x, this relation is symmetric.

    The relation is transitive if for all x, y, z is an element of X, if x~y and y~z, then x~z. It is also transitive because y=w.

    Could someone please show me where to go from here?
    Thank you very much
     
    Last edited: Mar 30, 2008
  2. jcsd
  3. Mar 30, 2008 #2

    quasar987

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    It is reflexive if for any (x,y) in R², (x,y)~(x,y).

    Is this really so?
     
  4. Mar 30, 2008 #3
    Is it not reflexive because (2,5) is not the same thing as (5,2)? I'm not really sure how to prove whether it is or it isn't. Could you please show me how?

    This one is reflexive, right? (x,y)~(w,z) if and only if x^2=w^2 because they are both squared. 2^2=2^2
    -2^2=2^2

    It is also symmetric and transitive, right? Could you show me how to prove it?(

    (x,y)~(w,z) if and only if xw=yz is not reflexive, right? because 2(5)=6(8) is not true

    Thank you very much
     
    Last edited: Mar 30, 2008
  5. Mar 31, 2008 #4

    HallsofIvy

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    No, this is not reflexive because (2,5)~(2,5) is wrong: (x,y)= (2,5) so y= 5. (w,z)= (2,5) so w= 2. 5 is not equal to 2 so y is not equal to w.

    Which one? If you are going to define a new relation, by "(x,y)~(w,z) if and only if x^2=w^2", please define it first, then ask the question! Yes, this is reflexive. No, it is not because of the squares but because x and w are both the first members of the pair. For any (x,y), (x,y)~(x,y) because x2= x2. (Notice that in that last equation I am using the fact that "=" itself is reflexive: a= a.)

    You prove something like that by showing that it satisfies the definition:
    A relation is symmetric if and only if a~b implies b~a. Suppose (x,y)~(w,z). Then x2= w2. But then, because w2= x2, we have (w,z)= (x,y). Do you see how I used the fact that "=" itself is symmetric?

    To show that this relation is transitive, we must show that "if (x,y)~(w,z) and (w,z)~(u,v) then (x,y)~(u,v)", the definition of "transitive". Okay, if (x,y)~(w,z) then x2= w2 and if (w,z)~(u,v) then w2= u2. From that (and the fact that "=" itself is transitive) we have x2= u2 which tells us that (x,y)~(u,v) and we are done.

    The "equivalence classes" are sets of point (x,y) where x2 is always the same. Since x2= w2 if x= w or x= -w, those are pairs of vertical lines, x= a and x= -a, on a graph.

    Where did you get "2(5)" and "6(8)"? The definition of reflexive is "a~a" for any a in the set. Here, your set is R2 or pairs of numbers. To show any relation on R2 is reflexive you must show (x,y)~(x,y) for any x and y. There are only two numbers involved, not 4. If (x,y)~(w,z) means xw= yz (first member times first member equals second member times second member), then (x,y)~(x,y) means x2= y2. That is not always true. (1, 2)~(1,2) is false because 12 is not equal to 22.

    You are welcome.
     
  6. Mar 31, 2008 #5
    Thank you very much

    (No, it is not because of the squares but because x and w are both the first members of the pair. For any (x,y), (x,y)~(x,y) because x2= x2. (Notice that in that last equation I am using the fact that "=" itself is reflexive: a= a.))

    I don't understand this part. In the previos example x and w were both the first members of the pair, but it wasn't reflexive. Could you please explain this to me?


    Now that I know that it is an equivalence class, the partition could be {{1}, {2}, {3}, {4}, {5},} right?

    Also, when you see (x,y)~(w,z) if and only if y=w and you are trying to prove whether or not it is reflexive, it means that if (x,y) for all x they are an element of X, right? If not, could you explain it to me, please?

    Thank you
     
    Last edited: Mar 31, 2008
  7. Mar 31, 2008 #6

    HallsofIvy

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    In the previous example, the relation was (x,y)~(w,v) if y= w. y and w are not both the first members of the pair (nor both the second members). In particular, (1, 2) is not "related" to itself in this relation because if (x,y)= (w,v)= (1, 2) y= 2 and w= 1 so y is NOT equal to w.

    No, no, no! This is an equivalence relation on R2 Equivalence classes are sets of pairs of numbers. Since x2[/sup] is always the same, Equivalence classes would be, as I said, sets of numbers where y can be anything but x is fixed: vertical lines.

    I have no idea what you mean by "if (x,y) for all x". Saying (x,y)~(w,v) means that the relation is defined on pairs of things- presumably pairs of numbers.
     
  8. Mar 31, 2008 #7
    Thank you



    I'm looking for the partition. Could you please tell me how you would find that?

    Thank you
     
  9. Apr 1, 2008 #8

    HallsofIvy

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    Each set in a partition using an equivalence relation (called "equivalence classes") consists of all members of the original set that are equivalent. Since in this particular problem (x,y)~(u,v) if and only if x= u, equivalence classes consist of pairs of numbers in which the pairs have the same first member: {(a, y)|y is any real number}. Geometrically, as I said before, the eqivalence classes are vertical lines.
     
  10. Apr 6, 2008 #9
    Thank you very much

    Regards
     
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