Prove that $\gcd(a,0)=|a|$ I wish to prove a property of the $\gcd$, namely: $\gcd(a,0)=|a|$.
Notice that for some $x,y$ we know that $\gcd(a,0)= ax+  0 \cdot y$ such that $\gcd(a,0)$ is a multiple of $a$. How do we make the step to |a|?
 A: 
Basic fact. If $a \ne 0$. The largest value that divides $a$ is $|a|$.  

This shouldn't surprise anyone.  $|a| =  \pm 1*a$ so $|a|$ divides $a$ and for any $x > |a|$ then if $x*r = a$ then $\pm x*r =|a|$ and $\pm r = \frac {|a|}x < \frac {|a|}{|a|} = 1$.  So $1 < r < -1$.  The only  way $r$ can be an integer is if $r = 0$ and $x*0 = 0 = a$.  But we stated that $a \ne 0$.
Allowing for negative factors is a bit of a pain but... for positive values this is obvious and negatives is jsut "special cases".
(The definition of $x$ divides $a$ is that there exists an integer $m$ so that $x*m = a$.)

Basic fact.  All numbers divide $0$.

This should surprise everyone the first time the see it but it makes perfect sense if you think about it.  $0 = x*0$ and $0$ is an integer so every number divides $0$.
So in plain english:

If $a\ne 0$ then $\gcd(a,0) = |a|$.

If $a \ne 0$ then $\gcd(a,0)$ is the largest value that divides both $a$ and $0$.  As all values divide $0$ this is just the largest value that divides $a$.  And that is $|a|$.
This only seems weird for the same reason "all values divide $0$" seems weird.  Once you see what this literally means it should be obvious.
Note, if you do not stipulate $a \ne 0$ the statement is not true.  $\gcd(0,0)$ must be undefined.  As all values divide $0$ there can't be any largest.
A: Here are also two more results that can be used to prove such claim:

$gcd(a,b) = gcd(a,b+ka), k \in \mathbb{Z}.$
$gcd(a,a) = a$

A: Since the gcd has to be positive and cannot be greater than $a$,  Notice $0<\gcd(a,0)\leq |a|$, however, $gcd(a,0) =|k a|$ so $0< |ka| \leq |a|$, which must mean that $k=1$, such that $\gcd(a,0)=|a|$.
