Can there be a Quantifier that isn't stated inside a predicate?

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Lets say there a question to define a sentence with predicates and quantifiers such as : "every two real numbers have an integer between them"

With predicate : P(x, y ) = " x and y have an integer between them"

Domain : all real numbers

Then would the final answer look like this: ∀x ∀y(P(x, y))

or like this:

∀x ∀y ∃z (P(x, y))

which means that there is an integer z in between x and y.

or would you add it to P(x,y) as P(x, y, z) = "x and y have an integer z in between them"

Also, apart from the answer, is it even possible to have a quanitfier like this : ∀x ∀y ∃z (P(x, y)) where the z isn't even inside P(x, y)?

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$\forall x \forall y \ P(x,y)$ is a valid way to express the desired statement. In fact, your predicate $P(x,y)$ can be rewritten to have $z$ in there as follows: $Q(x,y,z)$ could be defined to say that $\exists z \in \mathbb Z\ \left[ x \leq z \leq y \vee y \leq z \leq x \right]$. If you are only wanting to use $\exists z$ without specifying the domain of that existential quantifier, you could rephrase it as $\exists z\ \left(z \in \mathbb Z \wedge \left[ x \leq z \leq y \vee y \leq z \leq x \right] \right)$. At any rate, $\forall x \forall y\ P(x,y)$ is equivalent to $\forall x \forall y \exists z\ Q(x,y,z)$.

Now, if you simply write $\forall x \forall y \exists z\ P(x,y)$, the $z$ isn't playing a part at all in the given predicate since the predicate only mentions $x$ and $y$. Imagine a function $f$ of two variables and a predicate like $f(x,y) = 1$. It is "grammatically correct" to write $\forall x \forall y \exists z\ f(x,y) = 1$, but what is the $z$ doing? It is totally superfluous. See Quantifying a variable without using it in the quantified statement for more on that.