The statement $$\forall x \in \mathcal{R} \exists n \in \mathcal{N} (n>x)$$ Is true, but $$\exists n \in \mathcal{N} \forall x \in \mathcal{R} (n>x)$$ Is false.
Why does reversing the order of the quantifiers cause the first statement to become false?
On
The first statement means : for all real numbers, there exists a natural number (depending on $x$) such that $n > x$. (this is clear as we just round up $x$ and add $1$)
The second means: there is a natural number (depending on nothing), such that for all $x$ in the reals we have $n > x$. So all reals must lie below some $n$ (while $x+1$ is not).
On
To recap my comment: Reversing the order of the quantifiers creates an entirely different statement that has nothing to do with the first. Even though it's a small change notationally, it's a huge one semantically.
To expand on it:
$\forall x, \exists n$ means for every $x$, we can find some $n$ that works. The choice of $n$ may depend on what $x$ is.
$\exists n, \forall x$ means the same $n$ works for all possible $x$. This is significantly more restrictive and therefore is not necessarily true when the "$\forall x, \exists n$" version is true.
To summarize:
$\forall x, \exists n$ means $n$ can vary based on what $x$ is.
$\exists n, \forall x$ means one value of $n$ has to work for all values of $x$.
On
You notice the difference between One for All and All for One? https://en.wikipedia.org/wiki/Unus_pro_omnibus,_omnes_pro_uno
Read them aloud in English, not math. The important thing is the order $n$ and $x$ are chosen. In the first, $x$ is chosen first and it says that whatever $x$ is chosen you can find a larger $n$. $=\lceil x \rceil +1$ will work. In the second, $n$ is chosen first and has to be larger than any $x$ you can pick. This is false because you can choose $x=n+1$