If $\exists a,b \in \mathbb{N}$ then $\exists k,l \in \mathbb{Z}$ s.t. GCD(a,b)=ak+bl, $l,b\in \mathbb{Z}$

Question

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.

Answer 1

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

Answer 2

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 $d\mid a$.