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I would like to know, if this negation is correct, and if not, an explanation on what is wrong. Any help would be appreciated :)

Original: $$ \forall \epsilon > 0 \exists \delta > 0 \forall x\in R (|x|<\delta \Rightarrow |x^3|<\epsilon ) $$ My negation:

$$ \forall \epsilon > 0 \exists \delta > 0\exists x \in R(|x|<\delta \vee |x^3|\geq \epsilon) $$ $$ or $$ $$ \forall \epsilon > 0 \exists \delta > 0\exists x \in R(|x|\geq \delta \Rightarrow |x^3|\geq \epsilon) $$

The exercise said to negate and transform until there are no $\neg$'s in it.

(Sorry if my math english isn't perfect)


-Update-

Thank you for the explanations, I think I've got it. Do you think this is correct?

$$ \exists \epsilon > 0 \forall \delta >0\exists x\in \mathbb R (|x|<\delta \wedge |x^3| \geq \epsilon) $$

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1 Answer 1

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Well, the formula $\forall x.\phi(x)$ says 'for all $x$, $\ \phi(x)$ is true'. Its negation is 'not for all $x$ will $\phi(x)$ hold', i.e. $\exists x.\lnot\phi(x)$.

Similarly, the negation of "$\forall \varepsilon>0. $ blah-blah" $\ $ is $\ $ "$\exists\varepsilon>0. $ not blah-blah."

Start from the most nested subformula, now it is $(|x|<\delta\implies |x^3|<\varepsilon)$. This is of form $A\implies B$, it is equivalent to $\lnot A\lor B$, whose negation is $A\land\lnot B$, i.e. $\ (|x|<\delta\land |x^3|\ge \varepsilon)$.
[This is the case when the premise is satisfied but not the corollary.]

Then apply $3$ times the previous observations.

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  • $\begingroup$ I don't quite understand the paragraph "Similarly, the negation..." How is this similar to your first observation? I see the for all quantifier, but the ">" kind of throws me in this context. Could you explain that? $\endgroup$
    – Petra
    Nov 3, 2014 at 21:04
  • $\begingroup$ In this case the notation $\forall \epsilon > 0$ is really just another way of writing $\forall \epsilon \in R^+.$ Does that help? $\endgroup$
    – David K
    Nov 3, 2014 at 21:33
  • $\begingroup$ It does. :) I'll try the exercise again and will get back with the result. Thank you so much! $\endgroup$
    – Petra
    Nov 3, 2014 at 21:42

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