Attempt to prove that (a,b) = (a,b,a+b)

Question

I want to check the validity of my proof to the following question.

Prove that (a,b) = (a,b,a+b) and more generally (a,b) = (a,b,ax+by).

Note that (a,b) is the greatest common divisor of $a$ and $b$, $a$ and $b$ are integers.

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My Prove:

let $(a,b)$ = $g$ and $(a,b,a+b)$ = $g^*$. Since $g$|$a$, $g|b$ $\rightarrow$ $g|a+b$ implying that $g|g^*$. On the other hand, $g^*|a$, $g^*|b$, following that $g^*|g$.

From the previous argument, $g=\pm g^*$. However, $g,g^*$, by definition, cannot be negative, thereby, $g=g^*$ as required. The general case can be proved analogously.

Thanks in advance

Answer 1

This proof is fine.

Depending on the level of your audience*, you may want to be more explicit in the implication "$g|a$ and $g|b$ implies $g|a+b$". In any case, showing the details here would certainly make the "analogously" for the general case more believable; as it is currently written, it took me a minute to see exactly how the analogy worked.

(* If you are in a course, and you don't have other instructions, a good rule of thumb is: your audience consists of you-from-a-week-ago, the weakest student in the course, and the grader.)