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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.

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$\forall x \in \mathbb Z^+, \exists y \in \mathbb R$ 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 $xy=1$.

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  • $\begingroup$ Thanks amWhy...I understood first statement...In second statment you read y only one time then loop through all values of x? $\endgroup$
    – zuby
    Commented Sep 22, 2016 at 19:01
  • $\begingroup$ 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 $xy = 1$ $\endgroup$
    – amWhy
    Commented Sep 22, 2016 at 19:04
  • $\begingroup$ 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 $x=\frac 1\pi.$ $\endgroup$
    – amWhy
    Commented Sep 22, 2016 at 19:11
  • $\begingroup$ You're welcome, @zuby! $\endgroup$
    – amWhy
    Commented Sep 22, 2016 at 19:12
  • $\begingroup$ 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) $\endgroup$
    – zuby
    Commented Sep 25, 2016 at 14:18

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