Addition/Multiplication of Equivalence classes, Well defined

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Homework Help Overview

The discussion revolves around the well-definedness of addition and multiplication of equivalence classes defined on the set ℝ, where two elements are equivalent if their absolute values are equal. Participants are tasked with demonstrating that multiplication is well defined and providing an example to show that addition is not well defined.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the definitions of well-defined operations, questioning whether different representatives of the same equivalence class yield the same or different results. They discuss specific examples, such as using 1 and -1, and 1/2 and -1/2, to illustrate their points.

Discussion Status

There is an ongoing exploration of the definitions and implications of well-definedness in the context of addition and multiplication of equivalence classes. Some participants have provided examples and clarified their understanding, while others are still questioning their interpretations and the implications of their findings.

Contextual Notes

Participants are navigating the definitions and implications of operations on equivalence classes, with some uncertainty regarding the outcomes of their examples and the definitions themselves. There is a focus on ensuring that the operations yield consistent results across equivalent representatives.

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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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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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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Yes, what you said: well definedness means getting the same results when using equivalent elements.
 
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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.
 
Try, e.g., 1+1/2 and 1-1/2.
 
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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 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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Good job, you got it.
 
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Thank you for the assistance and confirmation of my thoughts.
This problem is now solved.
 
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Glad to have helped, good job.
 
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