TITLE: Show that $\gcd(a, 0)$ exists and equals $|a|$ for all $a$ in $\mathbb Z$
QUESTION [8 upvotes]: I came up with the proof in the paragraph below. My question is about how I expressed the proof, and about the first part of the question above.
For one, my proof seems to me very wordy compared to proofs in my textbook or shown by my professors, so I'd appreciate input on how to express it in a more formal way. Also, I haven't shown that $d$ (where $d = \gcd(a, 0)$) exists and I don't see how I'd do so.

PROOF: Suppose $d = \gcd(a, 0)$, where $d$ is an integer. Then $d \mid a$ and $d \mid 0$. As every integer divides $0$, $d$ will be the largest divider of $a$. The largest divider of any integer is itself. However, $a$ may be negative and $d$, by definition, is greater or equal than zero, so $d = |a|$.

I appreciate any answers. Thanks!

REPLY [3 votes]: For what it's worth, I would reword it thus:

Suppose $d = \gcd(a, 0)$, where $d$ is an integer and $a$ is a positive integer. Then $d \mid a$ and $d \mid 0$. As every nonzero integer divides $0$, $d$ will be the largest divisor of $a$. The largest divisor of any positive integer is that positive integer itself, and so $d = a$. Given that $\gcd(a, -a) = |a|$, it follows that if $a$ is allowed to be negative, then $d = |a|$.

Okay, so that has a greater word count than what you wrote, but hopefully it's clearer.
I'm not a math teacher nor any kind of professional mathematician, so feel free to take this with a grain of salt.