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I'm working on some released exams, in particular this one and I am stuck on 4B.

I know I need to prove three properties to show that it is a group:

  1. The operation is associativity
  2. There exists an identity
  3. There exists an inverse.

I can find a solution for 2 and 3 somewhat easily:

For 2:

Let $x$ be an element belonging to the equivalence class. Then $[x + 0] = [x]$, which implies $[0]$ (not actually zero itself, but zero's equivalence class) is the identity element.

For 3:

Let $[x]$ be an element belonging to the equivalence class. Then $[x + (-x)] = [0]$ which implies $[-x]$ is the inverse of [x].

I am having trouble with the last part, associativity. I'm virtually lost:

Let $x,y$ and $z$ be elements of arbitrary equivalence classes. Then:

$$[[x + y] + z] = ?$$

I don't know what to do from here. Any ideas? Also, when the question states:

Be sure to check that the operation is well defined.

I don't understand what I am supposed to be looking for. My first though is that there isn't a unique identity or inverse... But the more I stare at the problem, I believe there is.

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  • $\begingroup$ Well defined means that if $[x] = [x']$ and $[y] = [y']$, then $[x]+[y] = [x']+[y']$. This says that the addition rule is not dependent upon your choice of representative. Equivalently, it means that if $x\sim x'$ and $y\sim y'$, then $x+y\sim x'+y'$. (The group operation respects the equivalence relation.) $\endgroup$ Feb 9, 2015 at 1:31
  • $\begingroup$ @CameronWilliams: Ok, I guess that is there to help with the identity element confusion I initially had. That should be simple enough: $x-x' = n$, $y-y' = k$ $x + y - x' - y' = x - x' + y - y' = n + k$. Thanks for clarifying that. $\endgroup$
    – Dair
    Feb 9, 2015 at 1:38
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    $\begingroup$ I wrote a detailed example of what it means for an operation to be “well-defined” and what goes wrong when it isn't. This might help you understand why we care whether the operation is well-defined. $\endgroup$
    – MJD
    Feb 10, 2015 at 17:05

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Note that in your answer, $$[[x+y]+z]$$ does not make sense. To work it out you would have to begin with $$[x+y]+z\ ,$$ which is a set plus an individual number, which cannot be done. The associative law states that $$(\,[x]+[y]\,)+[z]=[x]+(\,[y]+[z]\,)\ :$$ if you carefully use the definition of addition, and make sure you do the steps in the right order, you should find this is not too hard to prove.


Re: well-defined, let's start with an example of something which is not well defined. Suppose you try to order your equivalence classes by defining $$[x]>[y]\quad\hbox{if and only if}\quad x>y\ .$$ Then we have, for example, $$[0.5]>[0.2]\quad\hbox{and}\quad [0.5]<[1.2]\ .$$ But this does not make sense, because the equivalence class $[1.2]$ is the same as $[0.2]$, so we have said that it is both less than and greater than $[0.5]$. We say that the order specified here is not well-defined: the point is that since $[0.2]=[1.2]$, anything sensible that we say about this class cannot depend on the specific numbers $0.2$ or $1.2$.

Looking at addition, we have for example $$[0.2]+[0.5]=[0.7]\ .$$ If we calculate $[1.2]+[0.5]$ it is really the same sum, so we should get the same answer. And indeed, $$[1.2]+[0.5]=[1.7]=[0.7]\ .$$ What you have to prove is that addition always works "properly" in this respect, that is, $$\hbox{if}\quad [x_1]=[x_2]\ \hbox{and}\ [y_1]=[y_2]\ ,\quad\hbox{then}\quad [x_1+y_1]=[x_2+y_2]\ .$$ Good luck!

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  • $\begingroup$ Ok, I think I get it. As for the second part, I talk about it with Cameron. As for the first part, interpreting the operation as you say I should I get: $([x] + [y]) + [z] = [x] + ([y] + [z])$ Which implies: $[x+y] + [z] = [x] + [y+z]$ Which implies: $[(x+y) + z] = [x + (y + z)]$ Which using additions associative-ness gives: $[x+y+z] = [x+y+z]$ Correct? $\endgroup$
    – Dair
    Feb 9, 2015 at 1:49
  • $\begingroup$ Correct except that the logic is backwards, you need to start with $[(x+y)+z]=[x+(y+z)]$ and finish with $([x]+[y])+[z]=[x]+([y]+[z])$. $\endgroup$
    – David
    Feb 9, 2015 at 2:33
  • $\begingroup$ Ok thanks! Yeah, I wrote it backward mostly because it is a comment and I wanted to keep it compact, but still illustrate the crux of my point. I have it better written out in my notebook. Thanks! $\endgroup$
    – Dair
    Feb 9, 2015 at 3:36

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