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By the dupe, the table implies $\,(14441,3565) = (3565,189) = \ldots = (28,21) = (21,7) = (7,0) = 7\ \ $

I understand that Euclid's algorithm on GCD is based on doing division via subtraction $x = qy + r$. I also understand that the process is keep expressing the quotient in terms of the remainder.
Example to find GCD of $14441$, $3563$:

x = q * y + r
14441 4 3565 189
3565 18 189 161
189 1 161 28
161 5 28 21
28 1 21 7
21 3 7 0

So the GCD is $7$.
So basically we try to divide the $2$ original numbers and then try to see if the remainder can express evenly $y$ and keep doing that recursively i.e. try to find the smallest number that divides the remainder.
But I am not sure I understand the intuition behind the idea. Why would that process lead to the smallest number that divides the original $x$?
I also read that a core idea is that $gcd(x,y) = gcd(y,r)$ but I didn't really get that part too.
Could someone please help?

Bill Dubuque
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Jim
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  • if $d$ divides $x$ and $y$, then $d$ divides $r=x-qy$ – J. W. Tanner Feb 09 '21 at 17:28
  • @J.W.Tanner: I know that if $d|x$ and $d|y$ then $d|(x - y)$ but I wasn't aware of what you mentioned and I am not clear how to follow the thought for the original question using that – Jim Feb 09 '21 at 17:30
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    The key idea is that, the gcd of $x$ and $y$ must also be a factor of $(x-y)$, $(x-2y)$, $\dots$, $(x-qy)=r$. As the gcd is a factor of $x$, $y$ and $r$, we can express it using any two of them. – kctong529 Feb 09 '21 at 17:38
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    if $d|x$ and $d|y$, then $d|x$ and $d|qy$, so $d|x-qy$ – J. W. Tanner Feb 09 '21 at 17:43
  • @PM2Ring: Yes that is the definition of GCD i.e. the greatest number that divides both $x$ and $y$. – Jim Feb 09 '21 at 18:51
  • @PM2Ring: That was bad phrasing from my side. It is the smallest number reached from that set of divisions but somehow it is the greatest divider – Jim Feb 09 '21 at 20:56
  • @PM2Ring: None of the decreasing numbers besides $7$ is a divisor of $14441$ though. So how do they retain the factors along the process exactly? – Jim Feb 09 '21 at 21:22
  • Let us continue this discussion in chat. – PM 2Ring Feb 09 '21 at 21:24
  • @PM2Ring: Replied there, I am sorry just saw your message – Jim Feb 09 '21 at 22:29
  • By the dupe, the table implies $,(14441,3565) = (3565,189) = \ldots = (28,21) = (21,7) = (7,0) = 7,,$ where we use the common notation $,(x,y):=\gcd(x,y)\ \ $ – Bill Dubuque Feb 10 '21 at 00:13
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    https://math.stackexchange.com/questions/3379695/why-does-the-euclidean-algorithm-for-finding-gcd-work/3379763#3379763 – Ethan Bolker Feb 10 '21 at 00:16
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    Since the remainder sequence is strictly decreasing it must eventually reach $0,$ (by $\Bbb N$ is well-ordered), so the last nonzero remainder (here $7)$ is the gcd. – Bill Dubuque Feb 10 '21 at 00:18

1 Answers1

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You might find it easier to consider just a single step:

For any two numbers $x$ and $y$ with $y<x$, consider $x$ and $x-y$.

Any divisor of $x$ and $y$ is also a divisor of $x-y$ and $y$. Similarly, any divisor of $x-y$ and $y$ is also a divisor of $x$ and $y$. Thus gcd$(x,y)=$ gcd$(x-y,y)$.

We have thus reduced the pair of numbers and can repeat this over and over again.

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    @Jim I would add to S. Dolan's answer that if you examine the chart in your query, you will notice two attributes that will always be present, regardless of the initial values of $x$ and $y$ : [1] The numbers in the left hand column are strictly decreasing [2] The bottom row of the chart will always have $r = 0.$ – user2661923 Feb 09 '21 at 17:51
  • @S. Dolan: "Thus $gcd(,)= gcd(−,)$" and where does $r$ come in the picture? – Jim Feb 09 '21 at 18:10
  • If x-y>y then the next step would be x-2y and y . Repeating this as necessary you get x-qy=r. –  Feb 09 '21 at 18:12
  • It seems to me that your answer more focuses on the algorithm, but I still don't get the intuitive idea behind it. I do understand the process – Jim Feb 09 '21 at 19:58
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    gcd$(x,y)=$ gcd$(x-y,y)$ looks like an intuitive idea to me. –  Feb 09 '21 at 20:22
  • @S.Dolan:So if we have two numbers $x$ and $y$ with $x > y$ we try to see if $y$ divides $x$ evenly. If so then $y$ is the gcd. This is intuitively understood. Now if it does not evenly divide it means we get a remainder $r>0$. I also can intuitively understand that we now can turn our attention to $r$ since if we find that $r$ evenly divides $y$ it is evenly divides $x$ but it is not clear to me at a high level why can't there be some other factor e.g. $z$ inside $y$ where $y>r$ that divides both $x$ and $y$ evenly? Or the intuition is only part of going over the algebraic equalities? – Jim Feb 14 '21 at 16:54
  • $r$ is $x$ minus a multiple of $y$. It would therefore be a multiple of your $z$. –  Feb 14 '21 at 17:07