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While it's easy to show that something isn't commutative, it's not so easy to prove that something is anything – at least for me!

If we take a relatively simple example: $(\mathbb{Z}_{n}$ , $\times_{n})\space$ s.t. $\space a*b\space =$ $\space\overline{a\times_{n}b}$

It's definitely a binary operation: defined for any $(a,b) \in \mathbb{Z}_{n}$ , the image of any input $(a,b)$ is unique i.e. well-defined, and it's closed in $\mathbb{Z}_{n}$ – but how do you prove that is commutative?

My thinking: For $\space\overline{ab}\space$ to be equal to $\space\overline{ba}\space$ it must be that $ab \equiv ba \space(mod \space n) \iff ab-ba = nq$.

Therefore, if $\space\times_{n}\space$ is commutative then $\space ab - ba$ must equal $\overline{0}$. As $\overline{a}$ and $\overline{b}$ are residual classes in $\mathbb{Z}_{n}$ they are integers, then $ab - ba = 0$ and the result of zero is a factor of $n$, therefore $(\mathbb{Z}_{n}$ , $\times_{n})$ is commutative.

But, more generally, what steps should we take to prove commutativity of a binary operation?

Benjamin R
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  • Commutativity is a property on two elements so involving three in your assumption doesn't make much sense. – Qudit Sep 10 '15 at 06:47
  • @Qudit Thanks, that was dumb! Lots of passion, not much talent... but trying hard as it is so beautiful and wondrous. – Benjamin R Sep 10 '15 at 06:52
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    That's not how you do a proof by contradiction. You're trying to prove that the property holds for all pairs of elements, so for contradiction you should assume that it doesn't work for just one specific pair. – Qudit Sep 10 '15 at 06:58
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    Also, I think you'd be better off using a direct proof for this case. – Qudit Sep 10 '15 at 06:59
  • I see no problem with editing it to reflect your current best attempt to solve it. – Qudit Sep 10 '15 at 07:10
  • Okay I had the answer for the specific example, I just needed to tie it up properly as my Professor said. – Benjamin R Sep 10 '15 at 23:44
  • It's been seven years since I wrote this. I did eventually get 'A' grades in my 200 and 300 level papers — not from talent, only sweat and the kind help of excellent lecturers at my small university campus. Since graduating, I have never had the ability to make use of anything substantial I learned in mathematics in my career as a developer, and have forgotten nearly everything, and that is so profoundly sad to me, since studying mathematics was the most joyful experience in schooling I ever had :(((((( – Benjamin R Oct 15 '22 at 15:38

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$a \times_n b$ is the remainder of $ab$ divided by $n$, and likewise $b \times_n a$ is the remainder of $ba$ divided by $n$. Since $ab = ba$, they also have the same remainder when divided by $n$, so $a \times_n b = b \times_n a$.

In general, you have to show that $a \star b = b \star a$ without making any assumption about $a$ and $b$ beyond the fact that they are elements of the set under consideration, so that your argument will apply equally to all the elements of the set.

fkraiem
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  • I had come to the same conclusion, can you see my updated question (kinda did it just before you posted your answer) to try to finish it into a proper proof. – Benjamin R Sep 10 '15 at 08:50