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When we define the ring $\mathbb{Z}/n\mathbb{Z}$, and we define addition as $[a] + [b] = [a + b]$. This makes sense as a definition as if you add anything of the form $a + nk$ with something of the form $b + np$ you get something of the form $(a+b) + nq$. On the other hand if you have something of the form $(a + b) + nq$, you can write it as the sum of something of the form $a + nk$ plus something of the form $b + np$.

So each class is contained in the other and therefore they are equal, that is, $[a]+[b]$ is a subset of $[a + b]$, and $[a + b]$ is a subset of $[a] + [b]$, where by $[a] + [b]$ I mean the set of all sums of something from $[a]$ with something from $[b]$.

But with multiplication, we define $[a][b]$ = $[ab]$, and it's true that if you multiply something of the form $a + nk$ with something of the form $b + np$ you get something of the form $ab + nq$, so $[a][b]$ is a subset of $[ab]$, but what about the reverse inclusion?

I'm struggling to justify that everything of the form $ab + nq$ can be written as a product of something from $[a]$ and something from $[b]$?

I understand that these definitions are well defined (they don't depend on our choices of representatives from the equivalence classes), I'm just struggling to justify the motivation behind the definition of multiplication.

Any help would be greatly appreciated! Thanks in advance!

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    $\begingroup$ To make sense of $[a][b]=[ab]$ you only need to comfirm that if $a',b'$ are different representatives of the classes $[a],[b]$ then you still get the same result. So $a'=a+nk$ and $b'=a+np$ and computation goes exactly as you did, and everything is fine. i.e. you are not trying to show that $[a][b]=[ab]$, but you are defining it that way. $\endgroup$ – user160738 Jan 22 '17 at 18:15
  • $\begingroup$ Ah yes, I see now, in order for it to make sense as a definition we need to check that taking the element [a] and the element [b] and multiplying them gives you the element [ab], and in order for us to do this we also need to check our choices of representatives don't matter. Thanks, it is clear now! $\endgroup$ – Jesse Parsons Jan 22 '17 at 18:22
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It is not true in general that any element of the form $ab+nq$ can be written as $(a+nk)(b+nm)=ab+n(kb+ma+nmk)$. Otherwise, it means that every integer can be written as $kb+ma+nmk$. If, for example, you choose $a,b,n$ such that all of them are divisible by some $1<p$, then $p\mid kb+ma+nmk$, so you cannot write 1 in this form.

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  • $\begingroup$ Ok that makes sense, thank you very much for your response. Upon further thought, I realise now that it doesn't need to be justified the way I was trying. $\endgroup$ – Jesse Parsons Jan 22 '17 at 18:20

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