why is $\exists z \forall x \forall y (x-y=z+2)$ false?

Question

In the rational numbers Q, is it true that: $$ \exists z \forall x \forall y \space (x-y=z+2) $$ This question showed up on my last test in discrete mathematics, and I have no idea why the right answer is that it is false.

Is it not possible to set $ z = x - y - 2 $ and conclude that such a $z$ will always exist in Q?

I am probably missing something pretty fundamental, can anyone help me understand this?

Answer 1

Your answer would be correct if the assertion was$$\forall x\forall y\exists z(x-y=z+2)$$because your $z$ depends upon $x$ and $y$. But the assertion$$\exists z\forall x\forall y(x-y=z+2)$$means that there is a rational number $z$ such that. for every rational $x$ and every rational $y$, $z+2=x+y$. And, clearly, no such number exists.

Answer 2

The assertion tells that $z$ is fixed. it doesn't depend neither of $x$ nor of $y$.

assume it exists.

then

$$1-1=z+2$$ and $$1-0=z+2$$

what is the conclusion.