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Can you guys help me prove or disprove this? Also this is what I got for the negation of this statement: There exists $x \in \mathbb{R}$ for all positive real numbers $\epsilon$ such that either $x > (-\epsilon)$ or $x < \epsilon$

Is my negation correct? Where should I start with my proof?

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    $\begingroup$ No, the negation is not correct. Negating an "or" produces an "and" (and vice versa). Start with $0$. $\endgroup$ – Daniel Fischer Sep 24 '13 at 23:27
  • $\begingroup$ What do you mean by start with 0? $\endgroup$ – Jake Park Sep 25 '13 at 0:01
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Your negation is not quite correct, because the negation of

either $x\le-\epsilon$ or $x\ge\epsilon$

is not

either $x>-\epsilon$ or $x<\epsilon$;

it’s

$x>-\epsilon$ and $x<\epsilon$.

However, the original statement is not true: it has one (and only one) exception. What real number is that exception?

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  • $\begingroup$ How do you know the original statement is not true? $\endgroup$ – Jake Park Sep 25 '13 at 0:07
  • $\begingroup$ @Jake: Because I know a counterexample. Note that the original statement is equivalent to saying that for each real number $x$ there is an $\epsilon>0$ such that $|x|\ge\epsilon$; there’s one real number for which that’s clearly false. $\endgroup$ – Brian M. Scott Sep 25 '13 at 0:09
  • $\begingroup$ Would that be 0? $\endgroup$ – Jake Park Sep 25 '13 at 0:51
  • $\begingroup$ @Jake: It would indeed. For any non-zero $x$ you can take $\epsilon=|x|$, but if $\epsilon>0$, it’s never true that $0\le-\epsilon$ or $0\ge\epsilon$. $\endgroup$ – Brian M. Scott Sep 25 '13 at 0:57
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Let $x \in R$.

Let $\epsilon \in R : 0 < \epsilon < x$ or $ \epsilon < -x < 0$.

Since $x + \epsilon \in R$, and $0 < \epsilon$, your statement is true.

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  • $\begingroup$ why $x>0$? surely $\exists x\in\mathbb{R}$ such that $x\not>0$... $\endgroup$ – Tim Sep 24 '13 at 23:47
  • $\begingroup$ And even x = 0, but I thought that was all too clear. $\endgroup$ – Don Larynx Sep 24 '13 at 23:57

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