Does $\gcd(a,b)=\gcd(\left|a\right|,\left|b\right|)$ imply that GCDs are nonnegative?

Question

This previous MSE question is related, but my present inquiry is not a duplicate of that one.

Here is my question:

Does $\gcd(a,b)=\gcd(\left|a\right|,\left|b\right|$) imply that GCDs are nonnegative?

MY ATTEMPT

I know from this MSE question that $\gcd(0,0)=0$, so that takes care of the case when $a = b = 0$.

However, I then have the following comment from gandalf61 underneath a related MSE question: "Because part of the definition of gcd is that it is a positive integer?" in response to the question "Prove that $\gcd(a,0)=\left|a\right|$."

Thus, I am all the more confused.

Answer 1

I think your question is defintional and doesn't have an exact answer. This is like asking to prove, $0^0=1$. We can't proof that, we simply take it as a definition, so that we can write, $$(x+1)^n = \sum_{k=0}^n \binom{n}{k}x^k$$ without worrying about specifying $x\ne 0$. We could choose, $0^0=0$ instead, but then we must have $x\ne 0$.

In your case, I prefer to follow Apostol's number theory, and define the gcd as the smallest positive integer in the set $$\{an+bm|n,m\in \mathbb{Z}\}$$ In other words the gcd is really the collection of all possible linear combinations, for convenience we pick a representative. This notion that the gcd forms a lattice, makes more sense latter on in other theorems in number theory too. The first theorem proved is that the gcd (being the smallest positive element in the set) does indeed divide every other element in the set above, meaning that the set of all linear combinations of any two numbers is precisely the same set as all the multiples of the gcd. Isn't that cool. Defining the gcd any other way wouldn't feel right to me.