Q:

HCF of two integers (a,b) =0

Suppose $hcf(p,q) = 0$ , is it even possible to prove that $p=q=0$? 
My answer to that is 
$x = hcf\left(\frac ph,\frac qh \right)$. We have to prove that $x=0$.
Since $x$ is a common of $\frac ph$ and $\frac qh$, there exists integers $m_a$ and $m_b$ such that
$$\frac ph=x∙(m)_a \qquad     and \qquad     \frac qh=x∙(m)_b$$ 
and multiplying everything by h gives us
$$p=hx∙m_a \qquad     and      \qquad q=hx∙m_b$$
Consequently, $hx$ is a common factor for $p$ and $q$.
We know however that $h$ is the highest common factor of $p$ and $q$ and so, any other common factor of $p$ and $q$ must also be a factor of $h$. In the end, $hx$ must be a factor of $h$, i.e. there exists such integer y such that h=(hx)∙y 
It follows immediately that $xy=0$ for which the only possible solution are $x=y=±0$. By convention, however, it doesn't matter which one to choose when computing the $hcf$ of those two prime numbers.  So, this gives us the answer $x=0$, i.e. hcf of $p$ and $q$ is $0$.
Is this explanation correct?  If not, how would you write it out? 

A:

I doubt it is possible to have the HCF of two integers equals 0.  
If p ≠ 0, the factors of p does not include 0.  So 0 must not be the HCF of p and q. 
Similarly, if q ≠ 0, the factors of q does not include 0.  So 0 must not be the HCF of p and q either.  
Now for the final case: p = 0 and q = 0.  If we adapt the definition of factors for integers (rather than natural number), then any integer can be the factor of 0.  So, we can always find an integer > 0 that is the HCF(0, 0).  So, HCF(0, 0) is undefined (an integer as large as you would like).  
Combining all the cases, HCF(p, q) can never be 0. 