Order of parameters in quantified predicates I'm studying up for my midterm in Discrete Math and I've been looking at sample questions and their solutions. There is one I don't really understand and I was hoping someone could help me out.
2. Let the domains of x and y be the set of all integers.
Compute the Boolean values of the following quantified predicates:

  All x, Exist y, (x^2 < y)
  Exist y, All x, (x^2 < y)
  Exist x, All y, (x^2 >= y)
  All y, Exist x, (x^2 >= y)

Solution:
  All x, Exist y, (x^2 < y) = T
  Exist y, All x, (x^2 < y) = F
  Exist x, All y, (x^2 >= y) = F 
  All y, Exist x, (x^2 >= y) = T

I'm not really sure if I'm understanding this or not. The first solution appears to say there exists one integer that is greater than every integer squared? I guess that makes sense on a per-integer basis, but the second solution appears to say the same thing, in a different order, but is false.
I know there's more here that I'm just not seeing. Would somebody mind explaining the nature of the problem and solutions? It would mean a lot, thanks!

$\forall x.P$ means that every possible value of $x$ will make $P$ true.
$\exists x.P$ means that there is a value of $x$ that will make $P$ true.
One such value is enough, but there has to be at least one.
Usually each variable is restricted to some domain. For example, since this is
a discrete math course, can we stipulate that $x$ and $y$ represent integers?
Now consider, the sentence, "For $x = 4,$ there is a value of $y$ that satisfies $x^2 < y.$"
Is that true?
How about, "For $x = 17,$ there is a value of $y$ that satisfies $x^2 < y$"?
In fact, we could set $x$ equal to any integer, and there would still be a value
of $y$ that satisfies $x^2 < y.$ So the statement, "There is a value of $y$ that satisfies $x^2 < y,$" which we can write as $\exists y.(x^2<y),$ is true for all $x.$
And that is what $\forall x.\exists y.(x^2<y)$ says.
Now consider the sentence, "If $y = 4,$ every possible integer $x$ satisfies $x^2 < y.$" 
Is it true?
How about, "If $y = 17,$ every possible integer $x$ satisfies $x^2 < y$" ?
In fact, is there any value to which we can set $y$ so that $\forall x.(x^2<y)$
(that is, so that every possible integer $x$ will satisfy $x^2 < y$)?
No, there is not. 
The statement $\exists y.\forall x.(x^2<y)$ asserts that there is such
a value of $y,$ so that statement is false.
