# 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?

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What if $x=0$? $0$ is an integer. – Brian M. Scott Jan 22 '13 at 4:16
Thank you! I'm still confused. 6*2 is equal to 12. so isn't this making the statement true already?? – hibc Jan 22 '13 at 4:20
It says for all x. It's not true for x=0, so the statement is false. – Ted Jan 22 '13 at 4:21
No, because the statement says that no matter what integer $x$ I pick, you can find an integer $y$ such that $xy=12$. You can do it if $x$ is $\pm1,\pm2,\pm3,\pm4,\pm6$, or $\pm 12$, but not otherwise. – Brian M. Scott Jan 22 '13 at 4:22
Thank you @BrianM.Scott. What if the order of quantifiers were changed. (∃y)(∀x)P(x,y) then would it be true (I think not all x is true for this case so..) ?? – hibc Jan 22 '13 at 4:26

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 $y$.

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Thank you very much :) – hibc Jan 22 '13 at 4:30
@hibc: You’re very welcome. – Brian M. Scott Jan 22 '13 at 4:31