gcd$(x,y) = 1$
how can i prove that there exist $m, n \in \mathbb{Z}: mx+ny = 1$
I'm studying abstract algebra - cyclic group part, and wondering how $<d>$ is same group with $<r,s>$.
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1https://artofproblemsolving.com/wiki/index.php/Bezout%27s_Lemma – 1123581321 Sep 27 '20 at 07:28
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The classical proof is obtained from the Euclidean Algorithm. – player3236 Sep 27 '20 at 07:36
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1See also https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Proof and https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm. – MvG Sep 27 '20 at 07:42
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Let be $A=\{mx+ny\in \Bbb N-\{0\} \, |\, m, n\in\Bbb Z\} $. Surely $A$ is a non-empty subset of $\Bbb N$, so it must have minimal element, call it $d$. We want to prove that $d=\gcd (x, y) $, to this end let's write $d=mx+ny$.
On the one hand surely if $d'$ is a common divisor of $x$ and $y$ it must be a divisor of $mx+ny=d$.
On the other hand $d$ must be a common divisor of $x$ and $y$: infact we can write $x=ad+r$ with $0\le r<d$ but so $x=a(mx+ny) +r$ so $r\in A$ but $r<d$ that implies $r=0$ and so $d|x$. In the same way we can prove that $d|y$.
Now we have that $d$ is one common divisor of $x$ and $y$ and such that every other divisor divides it, so it must be $d=\gcd (x, y) $
Alessandro Cigna
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I'm not sure why d is the biggest divisor. I guess i can only see d is common divisor. – hydthemoon Sep 27 '20 at 09:29
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@Jaehyun.Kim Every common divisor $,d',$ is smaller than $,d,$ by $,d'\mid d\Rightarrow d'\le d.,$ The proof shows that any common divisor having such a linear form is necessarily greatest. – Bill Dubuque Sep 27 '20 at 09:42
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