Nested Quantifiers and Equivalences I'm a little confused with nested quantifiers when it comes to comparing two statements. I have this for example:
Assume: $$ P(x,y,z) : x + y = z $$
$$\forall x \forall y \exists z : P(x,y,z)$$
The domain is real numbers for all variables. Now I know the above statement evaluates to true since adding any two real numbers, you'll always have a value $z$  for which it is true.
Now the confusion arises with the following:
$$\exists z \forall x \forall y : P(x,y,z)$$
The way I read this: "There is a real number $z$ for every real number $x$ and for every real  number $y$". This statement is False but don't see why... wouldn't there be a value for $z$ for which this is true? I'm stuck on how to interpret these. They sound so similar yet I cannot see the difference.
 A: The second statement $$\exists z \forall x \forall y \;P(x, y, z)$$ is false because there is no ONE $z \in \mathbb R$ for which $P(x, y, z)$ is true for every possible pair of reals $(x, y)$. 
I.e., there is no ONE $z \in \mathbb R$ such that $P(x, y, z)$ is true no matter what the values of $x$ and $y$.
Is there any $z$ for which $1 + 2 = z$ and $2 + 3 = z$? If there were such a $z$, we'd have $1+2 = 2 + 3$, which is absurd.
A: Think of it as a game between players named $\exists$ and $\forall$. You’re $\exists$, and your goal is to make the statement $x+y=z$ true; your opponent is $\forall$, and his goal is to make it false. You pick your real numbers in the order in which the quantifiers appear.
With the first statement you always win: no matter what real numbers $x$ and $y$ he picks, you can pick the number $x+y$ for your $z$, and $P(x,y,z)$ will be true.
Now look at $\exists z\forall x\forall y(x+y=z)$: in this version of the game you have to play first. You pick some real number $z$. Can your opponent $\forall$ choose $x$ and $y$ to make you lose? Absolutely, and in lots of ways. For instance, he can choose $x=z$ and $y=1$. 
A: $\exists z \forall x \forall y : P(x,y,z)$ is the same thing as saying:
$\text{There exists a $z$ such that for all numbers $x$ and $y$, $x+y=z$ is true.}$
or equivalently,
$\text{There exists a $z$ such that the sum of any two numbers ($\in \mathbb{R}$) is equal to $z$.}$
A: Does there exist a real number which equals the sum of any two numbers?
The intended meaning of the second statement in words can get read "there exists a real number z such that for any real number x, for any real number y, [(x+y)=z]."  In other words, all pairs of real numbers would have to sum to this real number z if it existed.
