Logic questions over set of real numbers

Question

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?

Answer 1

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 $\exists x\forall y\neq0(xy=1)$ means that “there is a real number $x$ such that, for each non-zero real number $y$, we have $xy=1$”.

Answer 2

We have that

$$∀x\quad ∃z\quad ∀y \quad \left(z = \frac{x+y}2\right)$$

is false indeed we can easily find a counterexample for $x_1,x_2$ and $y_1,y_2$ such that $x_1+y_1\neq x_2+y_2$

$$\frac{x_1+y_1}2\neq \frac{x_2+y_2}2$$

The order is very important indeed for example the following

$$∀x\quad ∀y \quad ∃z\quad \left(z = \frac{x+y}2\right)$$

is true.

The statement $$∃x\quad ∀y ≠ 0\quad \left(xy=1\right)$$

means exists $x$ such that for all $y$ not equal to $0$ we have $x$ times $y$ is equal to $1$.

Also in that case the statement is true and in those cases to show that we need a proof.

Answer 3

The key here is that $\forall x\exists y P(x,y)$ is different (meaning is different) from $\exists y \forall x P(x,y)$. So the order of the quantifiers is important and the statement needs to be read/interpreted accordingly (see other answers).

Answer 4

I assume that we can accept this rule:

$(∃u \in \mathbb R)(∀v\in \mathbb R)\,P(u,v) \, \implies \, (∀v\in \mathbb R)\, P(u_0,v)$

---

$(∀x \in \mathbb R)(∃z \in \mathbb R)(∀y\in \mathbb R)\,(z=\frac{x+y}{2}) \implies$
$ (∀x \in \{0\})(∃z \in \mathbb R)(∀y\in \mathbb R)\,(z=\frac{x+y}{2}) \implies$
$ (∃z \in \mathbb R)(∀y\in \mathbb R)\,(z=\frac{0+y}{2}) \implies$
$ (∀y\in \mathbb R)\,(z_0=\frac{y}{2}) \implies$
$ (∀y\in \mathbb R)\,(y = 2z_0) \implies$
$ (∀y\in \{2z_0 + 1\}) \, (y = 2z_0) \implies$
$ (2z_0 + 1 = 2z_0) \implies 1 = 0$

---

$∃x\, ∀y ≠ 0\, \left(xy=1\right) \implies ∀y ≠ 0\, \left(x_0y=1\right)$