# Truth value of quantified statement

I have $P(x,y) \iff xy=1$; the universe of discourse for $x$ is the set of positive integers, and the universe of discourse for $y$ is the set of real numbers. $$\forall x \exists y P(x,y)$$

I'm confused about reading the above statement. I'm quoting above statement from a book that says the statement read as "For every positive integer $x$ there is a real number $y$" such that $xy=1$. so the statement is true...I want to know what value, that makes this statement true, of $y$. is there only one value of $y$ for all values of $x$? or I need to change the value of $y$ every time when i change $x$.

Please If someone could explain it would be appreciated. thank you.

$\forall x \in \mathbb Z^+, \exists y \in \mathbb R$ such that $xy = 1.$ True. The value of y that makes the statement true will change according to the value of $x$: $(x, y): (2, \frac12), (8, \frac 18), ...$. we can take any integer $x$, and put $y = \frac 1x$ so that $xy = 1$ is true. In truth, we can say that every positive integer has a multiplicative inverse, y, in $\mathbb Q \subset \mathbb R$

The expression $\exists y \forall x, P(x, y)$ is false when $P(x, y)$ means $xy = 1$. That is, there does not exist a value of $y \in \mathbb R,$ such that for each and every positive integer , we have that $xy=1$.

• Thanks amWhy...I understood first statement...In second statment you read y only one time then loop through all values of x?
– zuby
Commented Sep 22, 2016 at 19:01
• Yes, indeed. There is one single y, and when we loop through values of x, we will soon realized that most, if not all, do not satisfy $xy = 1$ Commented Sep 22, 2016 at 19:04
• re, my last comment: Suppose $y = \pi \in \mathbb R$. Is every integer, x, satisfy the claim that $x\cdot \pi = 1$? No INTEGER x is such that $x=\frac 1\pi.$ Commented Sep 22, 2016 at 19:11
• You're welcome, @zuby! Commented Sep 22, 2016 at 19:12
• Please amWhy explain this statement also how to read?...I'm reading this statement like: first value of x for all value of y then x's 2nd value again for all value of y. is this right? ∀x∀yP(x,y)
– zuby
Commented Sep 25, 2016 at 14:18