Truth of statements about numbers Given the following statements:


*

*$\forall\, x,y \in \Bbb Q \quad \exists\, z \in \Bbb Q $ $\;$ $ : \left(x<z<y\right) \vee \left(x>z>y\right)$.

*$\forall \, x \in \Bbb R \quad \exists\,y\in \Bbb R : y^2= x$

*$\forall \, x \in \Bbb R^+ \quad \exists\,y\in \Bbb R : y^2= x$

*$\forall \, x \in \Bbb Z : | x | > 0$

*$\forall\, x,y \in \Bbb Q : \left(x<y \rightarrow \exists\, z \in \Bbb Q: x<z<y \right)$

*$\forall \, x \in \Bbb N \quad \exists\, y \in \Bbb N : x>y$

*$\forall \, x \in \Bbb R \quad \exists\, y \in \Bbb R : y^2=|x|$

*$\forall \, x \in \Bbb Z \quad \exists\, y \in \Bbb Z: x \lt y \lt x+1 $

*$\exists\,x \in \Bbb N \quad \forall\, y \in \Bbb N : x\le y$

*$\forall \, x,y \in \Bbb N \quad \exists \, z \in \Bbb N: x+z=y$

*$\forall\,x\in\Bbb R\quad\exists\,y\in\Bbb R:x\lt y \lt x+1$

*$\forall\,a,b\in\Bbb Q\quad\exists\,x\in\Bbb Q:ax=b$

*$\forall\,x\in\Bbb Z\quad\exists\,y\in\Bbb Z : x\gt y$

*$\forall\,x,y\in\Bbb Z\quad\exists\,z\in\Bbb Z:x+z=y$

*$\forall\,a,b\in\Bbb Z\quad\exists\,x\in\Bbb Z:ax=b$

*$\forall\,m\in\Bbb Z\quad\exists\,q\in\Bbb Q:m\lt q\lt m+1$
list which are true.
Only $2,6$ and $8$ are false, correct?
 A: The complete list of true statements is: 
3)  $\forall \, x \in \Bbb R^+ \quad \exists\,y\in \Bbb R : y^2= x$
5) $\forall\, x,y \in \Bbb Q : \left(x<y \rightarrow \exists\, z \in \Bbb Q: x<z<y \right)$
7) $\forall \, x \in \Bbb R \quad \exists\, y \in \Bbb R : y^2=|x|$
9)  $\exists\,x \in \Bbb N \quad \forall\, y \in \Bbb N : x\le y$
11) $\forall\,x\in\Bbb R\quad\exists\,y\in\Bbb R:x\lt y \lt x+1$
13) $\forall\,x\in\Bbb Z\quad\exists\,y\in\Bbb Z : x\gt y$
14.) $\forall\,x,y\in\Bbb Z\quad\exists\,z\in\Bbb Z:x+z=y$
16) $\forall\,m\in\Bbb Z\quad\exists\,q\in\Bbb Q:m\lt q\lt m+1$
For each the others (that are not true), you need only find a single counterexample in which the statement is false: $\color{red}{1, 2,4,6,8,10, 12, 15}$
Note that $1)$ is false whenever $x = y$.  It would be true, however, if we have $$\forall x, y \in \mathbb Q((x\neq y) \rightarrow \exists z \in \mathbb Q (x\lt z \lt y)\lor (x\gt z\gt y))$$
$(12)$ is false, if we take $a = 0, b= 2$, e.g., but is true if we state 
12) $\forall\,a,b\in\Bbb Q((a\neq 0)\rightarrow\exists\,x\in\Bbb Q(ax=b))$
A: Number 1 is false for x=3 and y=4 there doesn't exist a z. 
Number 4 as well for x=0.  
Number 10 for x=3 and y=2 there doesn't exist a z. 
Number 15 take a=2 and b=1 there doesn't exist x.
A: *

*False. Let $x:=0$ and $y:=0$.

*False. Let $x:=-1$.

*True. Let $y:=\sqrt{x}$.

*False. Let $x:=0$.

*True. Let $z:=\frac{x+y}{2}$.

*False. Let $x:=0$.

*True. Let $y:=\sqrt{|x|}$.

*False. Let $x:=0$.

*True. Let $x:=0$.

*False. Let $x:=1$ and $y:=0$.

*True. Let $y:=x+\frac{1}{2}$.

*False. Let $a:=0$ and $b:=1$.

*True. Let $y:=x-1$.

*True. Let $z:=y-x$.

*False. Let $a:=0$ and $b:=1$.

*True. Let $q:=m+\frac{1}{2}$.

A: 1 isn't true as I've pointed out in the comments, as $x=y$ gives a counterexample; 
2 of course isn't true; 
3 is true; 
4 is obviously wrong; 
5 is the correct version of 1, it's true; 
6 is obviously wrong; 
7 is true; 
8 is wrong; 
9 is true; 
10 is wrong; 
11 is right; 
12 is wrong, take $a=0, b \neq 0$; 
13 is true; 
14 is true; 
15 is wrong, there are the counterexamples of 12, and even more; 
16 is true; 
So the list of false statements is more than OP thought, it's actually 1, 2, 4, 6,8,10,12,15
