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How do you negate: $ \forall x\in \mathbb{R},\exists n,m\in \mathbb{Z} \hspace{3 mm}s.t.\hspace{1 mm}m ?

Here's what I got, but it's obviously flawed. $ \exists x \in \mathbb{R} \hspace {2 mm} s.t. \forall n,m \in \mathbb{Z} \hspace{1mm}, \neg (m < x \hspace{1.5mm}and\hspace{1.5mm} x and for $\neg (m < x \hspace{1.5mm}and\hspace{1.5mm} x

I got: $x\leq m \hspace{4mm} OR \hspace{4mm} n \leq x$

which is obviously not what the author of the book intended.

Another way for $\neg (m < x \hspace{1.5mm}and\hspace{1.5mm} x was:

$\forall m,n \in \mathbb{Z}, (m \leq x\hspace{2mm} and \hspace{2mm} n \leq x) or (m \geq x \hspace{2mm} and \hspace{2mm}n \geq x)$

but I'm not so sure about the alternative either. (it's somewhat intuitively right to me, but it might not)

Could you point out where I'm wrong, and how to fix that?

Thanks :D

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    The first one already gives you the contradiction you sought: for $\forall m (x \leq m)$ is impossible since it would mean that $x$ is minus infinity and also $\forall n (n \leq x)$ means $x$ is infinity.2012-03-07

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The first one already gives you the contradiction you sought: for $\forall m (x \leq m)$ is impossible since it would mean that $x$ is minus infinity and also $\forall n (n \leq x)$ means $x$ is infinity.