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Homework Help: Operation on Equivalent Classes

  1. Oct 13, 2012 #1


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    1. The problem statement, all variables and given/known data
    A solution to a problem has following operation:
    here, [(a,b)] and [(m,n)] are two equivalence classes.

    Is not

    Can anyone explain it to me?
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 13, 2012 #2


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    Science Advisor

    Not until you explain what you have written to us! You can create "equivalence classes" from any set of objects- and for almost all sets, "addition" is not even defined. Are you talking about equivalence classes of pairs of numbers? Integers, rational numbers, or real numbers? And what do you mean by the sum of two equivalence classes? That has to be defined separately. Even then you can create equivalence classes with different properties by using different equivalence relations.

    I think you mean the equivalence class, defined on the set of positive integers by "(a, b)~ (c, d) if and only if ad= bc" but then the sum of two equivalence classes is defined by the formula you give, you don't "prove" it. What is useful to prove in this case is that the "addition is well defined". That is, if (a, b) and (x, y) are in the same equivalence class and (m, n) and (p, q) are in the same equivalence class then (an+bm,bn) and (xq+ yp, nq) are in the same equivalence class.

    You can define "sum of equivalence classes" to be whatever you want as long as it is "well defined". This sum might be "well defined" for a different equivalence relation. What equivalence relation are you working with?

    I think you need to go back and review the basics of "equivalence relations" and "equivalence classes".
  4. Oct 13, 2012 #3


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    Thank you.
    That was somehow helpful. It is defined over the set of rational numbers for the relation you specified.
    Yes, I do not understand much and I find my notes/textbook insufficient. I searched online for some explanations, examples and did not find much information there.
    Still, thank You.
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