Given that  and  are positive integers with $\mid$, how would one prove that $\gcd(, ) = $? I'm having trouble showing that if  and  are positive integers with |, then (, ) = a.
I'm aware that the greatest common divisor of two numbers is the largest number that is a divisor of both numbers. Ex) $\gcd(4,2) = 2 $. But in this case $a=4$ and $b=2$ so the expression above would not work. I suppose what I'm asking is if variable $a$ is always the smallest of the two numbers given.
Note

I now realize that I should have given the problem some more thought before posting on here. I had been overlooking the | portion of the expression. Thank you for the help!
 A: Recall the definition of $\gcd$. Let $x,y \in \mathbb{Z}$. If $d = \gcd(x,y)$, then there exists integers $e,f$ such that $x = ed$ and $y = df$ where $d$ is the largest possible integer with that property. So turning to your question:
Let $a,b \in \mathbb{Z}$ such that $a|b$. Then there exists some $c \in \mathbb{Z}$, where $b = ac$. From the definition of $\gcd$ then, we have that $a = a \cdot 1$ and $b = ac$, so then $a = \gcd(a,b)$.
Notice, this is also due to the fact there cannot be any divisor of $a$ larger than $a$ itself. So we're showing that $a$ is indeed a divisor of both integers and in fact the greatest possible divisor of $a$ and hence the greatest possible divisor of both $a$ and $b$.
A: $\begin{array}{lll}\gcd(a,b) &=\min\{as+bt>0\mid (s,t)\in\mathbb Z^2\}\\
&=\min\{as+ant>0\mid (s,t)\in\mathbb Z^2\} & \text{since }a\mid b\iff b=na\\
&=a\times\min\{s+nt>0\mid (s,t)\in\mathbb Z^2\} & \text{use }\min(ax,ay)=a\min(x,y)\\
&=a\times \gcd(1,n) & \text{choosing }s=1-n,t=1\text{ gives }\gcd(1,n)=1\\
&=a\times 1=a\end{array}$
