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Addition/Multiplication of Equivalence classes, Well defined

  1. Jan 30, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    For the set ℝ, define ~ as x~y whenever |x| = |y|. Define the addition and multiplication of equivalence classes as:

    [x]+[y] = [x+y]
    [x]*[y] = [xy]

    a) Show that the multiplication of equivalence classes is well defined.
    b) Give an example that illustrates that the addition of Equivalence classes is not well defined.

    2. Relevant equations


    3. The attempt at a solution


    a) [x]*[y]=[xy]
    Well, let's focus on B for now.

    b) I don't understand this. How could it not be well defined? This is the definition of addition.
    I figure we MUST use the fact that x~y means |x| = |y|.
    So, I think of examples, such as x=1 y=-1 as equivalence relations.
    And then such examples as x=1/2 and y = -1/2

    So we see 1+(-1/2) would be equal to 1/2
    while -1 + (1/2) would be equal to -1/2

    1/2 does not equal -1/2 but by the definition of equivalence here |1/2| = |-1/2| so it checks out, right?
    Or am I wrong here and just need to show that 1/2 does not equal -1/2 so we have a contradiction?
     
  2. jcsd
  3. Jan 30, 2016 #2

    WWGD

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    Well-definedness means that the result is dependent (meaning different) for different representatives of the same class,

    i.e., that using different representatives will give you different results.
     
  4. Jan 30, 2016 #3

    RJLiberator

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    What you are saying is that
    Well defined means that if you use different representatives of the same class, you will get different results.

    Isn't it the opposite? If an operation is well defined and you are plugging in values from the same class, then you should get the same results. Am I wrong here?
     
  5. Jan 30, 2016 #4

    WWGD

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    Yes, what you said: well definedness means getting the same results when using equivalent elements.
     
  6. Jan 30, 2016 #5

    RJLiberator

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    Okay.
    So for part B, I need to show that in using different elements of the same classes, I arrive at different results thus addition here is not well defined.

    What I did in part b came down to the following:
    So we see 1+(-1/2) would be equal to 1/2
    while -1 + (1/2) would be equal to -1/2

    Here I reached the results of 1/2 and -1/2
    These are different results when we look at them, however, the stipulation of x~y whenever |x| = |y| states that they are the same.
    So my example, therefore, is no good?
    Or did I miss the point that showing that 1/2 and -1/2 are different means that it is not well defined.
     
  7. Jan 30, 2016 #6

    WWGD

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    Try, e.g., 1+1/2 and 1-1/2.
     
  8. Jan 30, 2016 #7

    RJLiberator

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    Absolutely, that shows 1.5 and then 0.5. Boom. Part B is mission accomplished.

    I actually think I thought about that same scenario on my run, which is why I rushed back in to solve it, but then forgot when I got up here so I posted this thread.
    Bravo.

    Now, for part A:

    This is more obvious to me in my head, but difficult to put down into math lingo.

    My thoughts:
    [x]*[y] = [xy]
    Since |xy| = |-xy|
    no matter what inputs you have for x and y this is going to work. The end solution doesn't depend on signs since it can be the negative of itself or positive and it is equivalent.

    Would it work if I just did it by brute force?

    Showing
    x*y = xy
    x*-y = -xy
    -x*y = -xy
    -y*-x = xy
    and saying xy ~ -xy ?

    I think that's the best I have right now.
     
  9. Jan 30, 2016 #8

    WWGD

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    Good job, you got it.
     
  10. Jan 30, 2016 #9

    RJLiberator

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    Thank you for the assistance and confirmation of my thoughts.
    This problem is now solved.
     
  11. Jan 30, 2016 #10

    WWGD

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    Glad to have helped, good job.
     
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