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

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

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 $$\neg (m < x \hspace{1.5mm}and\hspace{1.5mm} x<n)$$ 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 way you wrote it is the correct way. – user17090 Mar 7 '12 at 9:25
But the exercise was this: Prove that: If x is an arbitrary real number, then there are integers n and m such that m < x < n I was thinking about proving by contradiction (pointing out that if there's no m,n >= x, then x is a lower bound of the integer set) – user269334 Mar 7 '12 at 9:32
the first one didn't look so helpful for task...but anyway, thanks for the reply Ali :D – user269334 Mar 7 '12 at 9:34
There is a slight mistake though: in your negation you replace $\mathbb{Z}$ by $\mathbb{N}$ which you shouldn't do. – user17090 Mar 7 '12 at 10:04
I just fixed it : ) Anyway, is the second one also right??? – user269334 Mar 7 '12 at 10:15
up vote 0 down vote accepted

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.

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