I tried to search for duplicate questions but I did not find any.
So here is my initial question:
Is it true that $\gcd(\left|a\right|,\left|b\right|) \leq \min(\left|a\right|,\left|b\right|),$ if $ab \neq 0$?
MY ATTEMPT
Since $\gcd(a,b)$ is the greatest common divisor of $a$ and $b$, in particular, it is a common divisor of $a$ and $b$. Thus we have both $$\gcd(a,b) \mid a \implies \gcd(a,b) \leq \left|a\right|$$ and $$\gcd(a,b) \mid b \implies \gcd(a,b) \leq \left|b\right|,$$ where the conclusions $\gcd(a,b) \leq \left|a\right|$ and $\gcd(a,b) \leq \left|b\right|$ are justified by $ab \neq 0$. Thus, we obtain $$\gcd(a,b) = \gcd(\left|a\right|,\left|b\right|) \leq \min(\left|a\right|,\left|b\right|).$$
Here is my final question:
Is this proof correct? If not, how can it be mended so as to produce a valid argument?
For $x$ and $y$ integers, the implication $x\mid y \implies |x|\le |y|$ is only valid for $y\ne 0$.