nested quantifiers In the domain of integers, $P(x,y)$. predicate "$xy = 12$"
I'm not sure why
$(\forall x)(\exists y)P(x,y)$  is false statement.
"For all $x$, there are some $y$, such that  $xy = 12$".
ex.: $6\cdot 2 = 12$.   let $x$ be $6$ and $y$ be $2$.  
Isn't this true?  
 A: If $P(x,y)$ is the statement $xy=12$, then over the domain of integers the statement $\forall x\exists y P(x,y)$ says:

$\qquad\qquad\qquad\qquad$ for each integer $x$ there is an integer $y$ such that $xy=12$.

Informally this says that no matter what integer I pick for $x$, you can find an integer $y$ such that $xy=12$. From elementary arithmetic you know that this means that $y=\frac{12}x$. But $\frac{12}x$ isn’t always an integer even when $x$ is. In fact, it’s an integer if and only if $x$ is $\pm1,\pm2,\pm3,\pm4,\pm6$, or $\pm12$. If I give you $x=5$, for instance, the only $y$ that makes $xy$ equal to $12$ is $\frac{12}5$, which is not an integer. And if I give you $x=0$, your situation is truly hopeless: there isn’t even a real number $y$ such that $0\cdot y=12$.
The statement $\exists x\forall y P(x,y)$ means:

$\qquad\qquad\qquad$ there is some integer $x$ such that no matter what integer $y$ is, $xy=12$.

This is clearly false. No matter what $x$ you try, if $xn=12$ for some integer $n$, then $x(2n)=24\ne 12$, so it’s not true that $xy=12$ for every integer $y$.
