If $a\mid b$ and $a>0$ then $(a,b)=a$

Question

Now let $d=(a,b)$. So $d=ax+by$. Since $a\mid b$ , so we have $aq=b$, where $q \in \mathbb{Z}$.

so we have $d=ax+aqy$. since a divides R.H.S , so it must divide L.H.S. So $a\mid d$ Also $d\mid a$ as its GCD of $a$ and $b$. So $d=a$. I am not sure though about proof.

Answer 1

I would do like this:

Assume $d=(a,b)$ so $d\mid a$. On the other hand $a\mid b$ so we have $b=pa$. By Bézouts identity $$d=an+bm$$ Since $b=pa$ we get $$ d=an+pam $$ Therefore $a\mid d$ and $d\mid a$ which can only be true if $a=d$.

Answer 2

Definition. Let $g,a,$ and $b$ be positive integers. Then $g = (a,b)$ if and only if

1. $g|a$
2. $g|b$
3. If $h|a$ and $h|b$, then $h|g$.

If $a|b$, then $(a,b)=a$

Proof

1. a|a
2. a|b
3. If $h|a$ and $h|b$, then $h|a$.

Hence $(a,b)=a$.

Answer 3

Let $d=(a,b)$.

We have $a\mid a$ and $a\mid b$, which implies $a$ is a common divisor of both $a$ and $b$. Now $d$ is greatest common divisor of $a$ and $b$, so $a\leq d$.

Next $d\mid a$ and $a>0$ gives $d\leq a$.

Thus, $d=a$.