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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. $$ \exists x \in \mathbb{R} \hspace {2 mm} s.t. \forall n,m \in \mathbb{Z} \hspace{1mm}, \neg (m < x \hspace{1.5mm}and\hspace{1.5mm} x$$\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 way you wrote it is the correct way.2012-03-07
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    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)2012-03-07
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    the first one didn't look so helpful for task...but anyway, thanks for the reply Ali :D2012-03-07
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    There is a slight mistake though: in your negation you replace $\mathbb{Z}$ by $\mathbb{N}$ which you shouldn't do.2012-03-07
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    I just fixed it : ) Anyway, is the second one also right???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.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.