# Show that \gcd(a,b)=|a| \iff a | b?

Show that $gcd\left(a,b\right)=|a|⇔a\mid b$?
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gunithd5
Step 1
Assume that . I use the following definition of $D=gcd\left(a,b\right)$:
1) D is a common divisor of a and b;
2) Every integer $d\in \mathbb{N}$ which is a common divisor of a and b divides D.
Proposition 1: If $gcd\left(a,b\right)=a$ then $a\mid b$
Proposition 2: If $a\mid b$, then $gcd\left(a,b\right)=a$
Proof: Since $a\mid a$ and $a\mid b$, then any integer d such that $d\mid a$ and $d\mid b$ satisfies also the condition d|a.
From and $d\mid a$ we conclude that $gcd\left(a,b\right)=a$
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ophelialee4xn
Step 1
It is true that the final assertion holds if $s=0$ and $t=1$, but it is not true that that $as+bt=az=b$ can only hold if $s=0$ and $t=1$ is false. Take, for example, $a=2$ and $b=4$. Then $\left(-5\right)a+3\left(b\right)=b$.
Remember that, by definition, $gcd\left(x,y\right)\mid x$ and $gcd\left(x,y\right)\mid y$. That gives one implication.
For the converse, remember that d|x and d|y implies $d\mid gcd\left(x,y\right)$.