Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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 $y$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.