Quick proof with Euclid's Algorithm

Question

I'm a little stuck on an exercise in a textbook of mine, and need some help.

Let

$J = \{e \in \mathbb{N}^* | e = ar + bs\}$ for some integers $r,s$.

and let $d$ be the least element of J. Show that $d | a$ and $d|b$.

I see that I'm supposed to divide d into a (stated in the back of the book) but I don't know where to go from there. If someone could lead me in the right direction, that'd be great (and explain step-by-step, cause I have trouble following sometimes.)

Thanks!

Answer 1

Let's write $d=as+br$ and let's suppose that $d$ does not divide $a$. Then you can write the Euclidian division of $a$ by $d$, $$a=dq+p$$

with $0 < p < d$ because $d$ does not divide $a$. But then, $p=a-dq=a-(as+br)q=a(1-s)+b(-rq)$ belongs to $J$ : this contradicts the definition of $d$ as the least element of $J$.

You can do the same if $d$ does not divide $b$.