is $\Bbb{Q}$ a GCD Domain? Wikipedia says every UFD is a GCD domain. As every field is a UFD so must be GCD domain.
Therefore in rationals, gcd(2,7) must exist. But what is it? I can divide both of them by any rational number. Like 10 divides both of them and so does 20 in rational domain.
 A: "The" gcd is in general not unique; it's only defined up to multiplication by units. But in a field, every nonzero element is a unit. So in $\mathbb{Q}$, every nonzero number is a gcd of 2 and 7.
A: In a commutative ring $R$ with 1: Let $g,x,y \in R$ and if $g$ is a gcd of $x$ and $y$ and $e \in R$ is a unit (so there exists $e' \in R$ with $ee' = 1$) then $ge$ is also a gcd of $x$ and $y$:
$g | x$ so $x = gx'$ for some $x' \in R$. So $x = g1x' = (ge)(e'x')$ so $ge | x$ as well. Similarly, $ge | y$ as well. 
And if $d|x$ and $d|y$ we know that $g | d$ (by the gcd property), so $d = gd' = (ge)(e'd')$ so $ge | d$ as well. So $ge$ also has the gcd property. This all follows from $x | y$ iff $xe | y$ for any $x,y$ and any unit $e$, in fact.
This shows that multiplying a gcd of $x$ and $y$ with any unit (i.e. invertible element) still gives a gcd, which is thus far from unique in general. I.e. in $\mathbb{Z}$ the only units are $1,-1$ so there the gcd is determined up to sign. But in $\mathbb{Q}$, where all non-zero elements are a unit, $\gcd(2,7) = x$ for any $x \in \mathbb{Q}\setminus\{0\}$ is a true statement. So people don't talk much about gcd's in this setting (thought they are defined, but pointless).
