Show that $\gcd(a,b)=|a| \iff a | b$ I'm reading A Computational Introduction to Number Theory and Algebra, which can be found here as a free download. From the book's exercises, I'm stuck with a proof to show that $\gcd(a,b)=|a| \iff a | b$.
My current reasoning goes as follows: since $\gcd(a,b) = |a|$, we see that $a\mathbb{Z} + b\mathbb{Z}= |a|\mathbb{Z}$.
Since $a|b \iff az = b$, for some $z \in \mathbb{Z}$, we see that both $az, b \in |a|\mathbb{Z}$.
So there exist $s, t \in \mathbb{Z}$ such that $as + bt = az = b$, which is true for $s = 0$ and $t = 1$.
Q.E.D.?
I guess that in the end my reasoning is wrong, but I can't explain it.

Assume that $a,b\in\mathbb{N}$. I use the following definition of $D=\gcd (a,b)$:
(i) $D$ is a common divisor of $a$ and $b$;
(ii) Every integer $d\in\mathbb{N}$ which is a common divisor of $a$ and $b$ divides $D$.
Proposition 1: If $\gcd (a,b)=a$, then $a|b$. 
Proposition 2: If $a|b$, then $\gcd (a,b)=a$. 
Proof: Since $a|a$ and $a|b$, then any integer $d$ such that $d|a$ and $d|b$
satisfies also the condition $d|a$. 
From $a|a$, $a|b$ and $d|a$ we conclude that $\gcd (a,b)=a$.
