Need to negate this proposition! I don't understand why the negation of: for all $n$ contained in the natural set and for all $m$ contained in the natural set there exists $x$ that is contained in $\mathbb R$ s.t $-n<x<m$ is:
there exists $n$ contained in the natural set and there exists $m$ contained in the natural set such that for all $x$ contained in $\mathbb R$ s.t one has either x<-n or x>m.
Why hasn't the and in the original text changed to or and why is $x$ not less than equal to $-n$ and more than equal to $m$? I thought you always replace and by or and vice versa and less than or more than by less than equal to or more than equal to.
 A: There is no logical and (i.e. $\land$) between the quantors in the formalized notation
$$\tag 1\forall n\in\mathbb N\colon\forall m\in\mathbb N\colon \exists x\in\mathbb R\colon (-n<x\land x<m). $$
What you pronounce as and in English is really just another such that, which would merely sound bad for linguistic reasons. So the negation becomes 
$$\tag 2\exists  n\in\mathbb N\colon\exists  m\in\mathbb N\colon \forall x\in\mathbb R\colon (-n\ge x\lor x\ge m). $$
It is not valid to replace $\ge$ with $>$ here to write 
$$\tag {2'}\exists  n\in\mathbb N\colon\exists  m\in\mathbb N\colon \forall x\in\mathbb R\colon (-n> x\lor x> m). $$
as that would make the statement stronger. However, as $(1)$ is true, $(2)$ is false and hence the stronger version is just as false.
But beware: Some folks use $\mathbb N=\{0,1,2,\ldots\}$ instead of $\mathbb N=\{1,2,3\ldots\}$. For these $(1)$ is false and $(2)$ is true  and $(2')$ is false. Thus in this interpretation $(2')$ cannot be the negation of $(1)$!
