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Equivalence Relation

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let X be Z*Z, i.e. X is the set of all ordered pairs of the form (x; y) with (x, y) are integers.
    De fine the relation R on X as follows:
    (x1^2, x2^2)R(y1^2, y2^2) = (x1^2 + x2^2) = (y1^2 + y2^2)

    2. Relevant equations
    By definition, an equivalence relation bears the following characteristics,
    Further information here, http://www.math.csusb.edu/notes/rel/node3.html

    3. The attempt at a solution
    Not an equivalence relation?
    Although it is reflective, the transitive and symmetric characteristics don't hold because if per say, we have x1 = 1, x2 = 2, y1 = 3, y2 = 4, the relation doesn't hold to start with...
    Is this a strong explanation? Or better suggestions?
    I feel that it's more complicated than this?
  2. jcsd
  3. Feb 21, 2010 #2


    Staff: Mentor

    You need to take pairs of numbers that satisfy the relation R. For example, (3, 4) and (2, sqrt(21)) satisfy the relation.

    One way to look at this is that two different points satisfy this relation if they are points on the same circle, where the circle is centered at the origin.

    Start with pairs of points points that satisfy the relation and then determine whether the characteristics hold. If they do, the relation is an equivalence relation; if not, it isn't.
  4. Feb 21, 2010 #3
    But does your example fit the relation? Don't the variables have to be integers?

    Going by your definition, if I do find a few cases that fit the relation and a few that don't, does that still mean the relation is an equivalence relation? Therefore, if let's say all variables are the same, ie x, y, z (for transitive) = 1, then the properties of an equivalence relation hold...?
  5. Feb 21, 2010 #4


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    Did you mean to write [itex](x_1^2,x_2^2)R(y_1^2,y_2^2)[/itex] instead of [itex](x_1,x_2)R(y_1,y_2)[/itex]? It just seems kind of weird.

    To show symmetry, you want to prove: if [itex](x_1,x_2)R(y_1,y_2)[/itex], then [itex](y_1,y_2)R(x_1,x_2)[/itex]. So by assumption, you have to start with two pairs that are related.

    The fact that (1,2) isn't related to (3,4) doesn't matter. Think about it. If you say that because you can find two elements that are not related, R is not an equivalence relation, you're saying that for R to be an equivalence relation, every element must be related to every other element. If that were true, there'd be really no reason to prove the individual conditions because they'd all trivially be true.
  6. Feb 22, 2010 #5


    Staff: Mentor

    You're right. I forgot that this was a relation on ZxZ. My main point still holds, though. For example, (3,4) and (-3, -4) would still be in the same class.
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