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I don't really understand the proof for this.
It says "consider a set $A=\{ax+by\mid x,y\in \mathbb{Z}\}$. Then let the smallest element of of $A$ be $d$.Then $d=ak+bn, k,n\in Z$. Show $d|a$ and $d|b$ so that $d$ is the common divisor. Then show that if $m|a$ and $m|b$ then $m \le d$."

I don't really know how if $d=ak+bn, k,n\in Z$ then $d$ divides $a$ and $b$ also I know how $m$ dividing $a$ and $b$ makes it less than $d$. I can understand it being equal but not less than.

Vinyl_cape_jawa
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Gooby
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    I have some issues understanding which parts you don't understand. Could you edit them as separate questions not one long sentence stating what you do understand and what you do not? – Vinyl_cape_jawa Oct 10 '19 at 09:55
  • Also, what are you allowed to say regarding $a\leq b$? Can you deduce it from the statement $a|b$ and a,b being natural numbers or do you need to give greater justification in your course? – Isky Mathews Oct 10 '19 at 10:31
  • I think it's fair to say you can assume one is larger than the other – Gooby Oct 10 '19 at 10:39
  • You can find the standard proofs in the linked dupe. If something is not clear then please ask questions in comments there. – Bill Dubuque Oct 10 '19 at 13:39

2 Answers2

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For the first part, do Euclidean division of $a$ by $d$: get $a=qd+r,0\le r\lt d$. But then, $r=a-qd=a-q(ak+bn)=a(1-qk)+bn\in A\implies r=0$, by minimality of $d$.

So $d\mid a$.

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$m|a$ and $m|b$ implies $m|(ak+bn)=d$ and hence $m \leq d$.

Kavi Rama Murthy
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