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My proof that addition in $G =\mathbb{Z}/n\mathbb{Z}$ is associative is as follows:

Proof：Before taking (mod $n$), we add members of $G$ in the same way as members of $H=(\mathbb{Z}, +)$ where we know that addition in $H$ is associative. Since taking (mod n) does not affect this addition, the addition in $G$ is also associative.

Does this proof seem like enough? I think I'm right but I often risk falling short in my proofs at times.

Shaun
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NoodleNami
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    That seems to take a lot for granted. For one thing, "members of $H$" are not the same as "members of $G$". For another, even if you pick a correspondence, you don't really add them the same way. For instance, if you take $n=2$, then adding the element "$1$" in $H$ to itself, you get the element $2$. But adding the elements in $G$ you get $0$, which is not equal to $2$. So it's not "the same". If you are being asked to prove this, I must guess that you do not have not yet established the framework that makes your statements both sensibly iterpretable and correct. – Arturo Magidin Feb 23 '21 at 22:52
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    Better proof?

    Since [a] in G is actually the members of the form a+kn where k is an integer, (similarly for [b] and [c] where we use l and m instead of k) we can show that ([a] + [b]) + [c] is a+b+c + (k+l+m)n = [a]+([b]+[c])

     – NoodleNami Feb 23 '21 at 23:03
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    Well, not really; now you seem to be treating $[a]$ as a single integer $a+kn$, etc. I mean, you can make this more formal, but it takes more than just waving your hands vigorously at equations. If you have proven that $[a]+[b]= [a+b]$ and that this is well defined, then a better way would be to write, explicitly, that $$([a]+[b])+[c] = [a+b]+[c] = [(a+b)+c] = [a+(b+c)] = [a]+[b+c] = [a]+([b]+[c]),$$ where the equalities are justified, respectively, as (i) by definition; (ii) by definition; (iii) by associativity in $H$ and well-definedness; (iv) by def.; (v) by definition. – Arturo Magidin Feb 23 '21 at 23:07
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    Before "taking ($\mod n$)".... is that what we do when we consider a residue in $G$? We assume there was some "real" integer and we somehow "mod" it? That's not the correct way of stating or thinking about it. And we don't "add" them the same way. For instance, when we add the numbers $13 + 54$ has nothing whatsoever to do with adding the number $49 +18$ even though they are the same thing $\mod 12$. – fleablood Feb 23 '21 at 23:39

2 Answers2

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As pointed out by @ArturoMagidin, your attempt assumes too much.

Here is a proof.

Let $[a]_n, [b]_n, [c]_n\in \Bbb Z/n\Bbb Z$. Then, by definition,

$$[a]_n=\{x\in \Bbb Z: n\mid (a-x)\}\quad\text{and}\quad [a]_n+_n[b]_n=[a+b]_n.$$

(I'll leave it to you to show that $+_n$ is well-defined.)

Now

$$\begin{align} [a]_n+_n([b]_n+_n[c]_n)&=[a]_n+_n[b+c]_n\text{ (by definition of }+_n\text{)}\\ &=[a+(b+c)]_n\text{ (by definition of }+_n\text{)}\\ &=[(a+b)+c]_n\text{ (by associativity of }+\text{)}\\ &=[a+b]_n+_n[c]_n\text{ (by definition of }+_n\text{)}\\ &=([a]_n+_n[b]_n)+[c]_n\text{ (by definition of }+_n\text{)}. \end{align}$$

Shaun
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Before taking (mod n), we add members of G in the same way as members of H=(Z,+)

This definitely needs to be proven. It's a breezy statement and as stated isn't clear what you even mean.

We DO need to state and DEFINE what if $[a],[b] \in \mathbb Z/n\mathbb Z$ what $[a]+[b]$ means.

And defining it is not as easy as it looks. $[a] = \{k\in \mathbb N: n|a-k\}$ so we can define $[a] + [b]:= [a+b]$ OR we could define it as $[a]+[b]=\{m+n|m\in [a];n\in [b]\}$ BUT 1) we have no idea whether $[a]+[b]:=[a+b]$ is well defined; we don't know that if $\alpha\in [a]$ and $\beta \in [b]$ that $\alpha + \beta \in [a+b]$ and 2) we don't have any reason to believe that $\{m+n|m\in [a];n\in [b]\}$ as a set is an equivalence class in $\mathbb Z/n\mathbb Z$.

However if we can prove that if $\alpha \in [a]=\{k\in \mathbb N: n|a-k\}$ and $\beta\in [b] = \{k\in \mathbb N: n|a-k\}$ then $\alpha+\beta\in [a]+[b]$, then the definition $[a]+[b] :=[a+b]$ will be well defined.

Pf: $\alpha\in [a];\beta\in [b]\implies n|a-\alpha;n|b-\beta\implies n|(a-\alpha)+(b-\beta)=(a+b) - (\alpha + \beta)\implies \alpha + \beta\in [a+b]$.

....

NOW you can use your one line breezy argument.

$([a] + [b]) +[c] = ([a+b])+c = [(a+b) + c]= [a+(b+c)] = [a] + [b+c]=[a]+([b]+[c])$.

fleablood
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