Addition/Multiplication of Equivalence classes, Well defined

In summary, the multiplication of equivalence classes is well defined, but the addition of equivalence classes is not well defined.
  • #1
RJLiberator
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Homework Statement



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.

Homework Equations

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?
 
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  • #2
RJLiberator said:

Homework Statement



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.

Homework Equations

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?

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.
 
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  • #3
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?
 
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  • #4
Yes, what you said: well definedness means getting the same results when using equivalent elements.
 
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  • #5
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.
 
  • #6
Try, e.g., 1+1/2 and 1-1/2.
 
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  • #7
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 into 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.
 
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  • #8
Good job, you got it.
 
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  • #9
Thank you for the assistance and confirmation of my thoughts.
This problem is now solved.
 
  • #10
Glad to have helped, good job.
 
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1. What are equivalence classes?

Equivalence classes are sets of elements that are considered equivalent based on a certain relation or equivalence relation. Elements within the same equivalence class are considered equivalent, and elements in different equivalence classes are considered not equivalent.

2. How is addition of equivalence classes defined?

Addition of equivalence classes is defined as the operation of combining two equivalence classes to create a new equivalence class. This is done by adding the corresponding elements from each class, and the resulting class contains all possible sums of the elements from the two original classes.

3. How is multiplication of equivalence classes defined?

Multiplication of equivalence classes is defined as the operation of combining two equivalence classes to create a new equivalence class. This is done by multiplying the corresponding elements from each class, and the resulting class contains all possible products of the elements from the two original classes.

4. What does it mean for an operation on equivalence classes to be well-defined?

An operation on equivalence classes is well-defined if it produces the same result regardless of which elements are chosen to represent the classes. In other words, the operation is independent of the specific elements within the classes, and only depends on the equivalence relation between them.

5. How are equivalence classes used in mathematics?

Equivalence classes are used in mathematics to group together elements that are considered equivalent and to simplify calculations. They are particularly useful in abstract algebra, where they are used to define mathematical structures such as groups, rings, and fields.

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