Logic questions over set of real numbers Why is $∀x\,∃z\, ∀y \, \left(z = \frac{x+y}2\right)$ false? Can't you have any real number for x and y, and have some real number z such that the equation is satisfied? What is a case where this is not true? I think that it bing false has something to do with the order of the quantifiers, but I cannot figure out what is amiss. Also, what does this: $∃x\, ∀y ≠ 0\, \left(xy=1\right)$ even mean? My translation would be: "there exists some $x$, where for all $y$ it is not the case that $xy = 1$ is false." Is that accurate?

The assertion $\forall x\exists z\forall y\left(z=\frac{x+y}2\right)$ means that for **each** real number $x$ **there is** a real number $z$ such that, **for each** real number $y$ you have $z=\frac{x+y}2$. This is clearly false. Take $x=0$, for instance. is there a real number $z$ such that $z=\frac y2$ for **each** real number $y$? Clearly not.

The order of the quantifiers matters. A lot.

The assertion $\exists x\forall y\
eq0(xy=1)$ means that " **there is** a real number $x$ such that, for **each** non-zero real number $y$, we have $xy=1$".