Euclidean proof

[Trashtalk217]

I thought it may be helpfull to provide a proof with the euclidean algorithm, since it is not directly obvious why the algorithm computes the greatest common divisor.

I don't know how the attribution, but I'll probably just give full credit to James Schloss, to avoid any confusion.

[june128]

Regarding the attribution: If you add this as a comment to this PR (replacing AUTHOR with your name), you can add something like the following to the license section of contents/euclidean_algorithm/euclidean_algorithm.md:

##### Text
The text of this chapter was written by [James Schloss](https://github.com/leios) and is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International License](https://creativecommons.org/licenses/by-sa/4.0/legalcode).
+ The proof was written by [YOUR NAME](LINK TO YOUR GITHUB/WEBSITE/WHATEVER) and is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International License](https://creativecommons.org/licenses/by-sa/4.0/legalcode).

[Trashtalk217]

I'd rather just give the rights to James since I don't want my real name in the book (yet) and I feel like writing "written by Trashtalk217" is a bit unprofessional.

[june128]

I'd rather just give the rights to James since I don't want my real name in the book (yet) and I feel like writing "written by Trashtalk217" is a bit unprofessional.

Fair enough. We can always change it in the future after all :)

[leios]

Thanks for the submission, we should certainly be adding proofs!

Should we provide a common format for proofs in the case that future proofs are added? (For example, should all proofs look like this: http://cheng.staff.shef.ac.uk/proofguide/proofguide.pdf).

I need to look at this proof a bit more rigorously and decide what we need for this section, in particular.

[dovisutu]

Yup, there should be a proof format. like this?

$$
\begin{align}
& \forall a,b \in \mathbb{N}, &  \quad \exists n \implies n & = (a,b) \\
& \therefore                  &                           n & \mid a, \; n \mid b\\
& \therefore                  &                           n & \mid a-b\\
& \therefore                  &                     (a-b,a) & = n
\end{align}
$$

[Trashtalk217]

Honestly I don't feel comfortable writing easy to read proofs yet. I still think proofs in the algorithm archive are a neat idea, but I don't think I'm qualified to write them. Maybe later.

Maybe just use @dovisutu 's proof, that could work.

[Amaras]

@leios, @Trashtalk217 How do we still want to have the proof in the AAA?

[leios]

My issue here is that I suck at proofs. If someone else can look at this and say it is valid and easy to read, I am happy to merge!

[dovisutu]

Yeah, it feels strange even when I am reading it, and that's written long ago... I don't think I'd rewrite this though, for I really can't get the time to do so. So maybe someone can write a more viable version than mine. :D