# Is this quantifier negation correct?

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?

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

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. $\ (|x|<\delta\land |x^3|\ge \varepsilon)$.
Then apply $3$ times the previous observations.
• In this case the notation $\forall \epsilon > 0$ is really just another way of writing $\forall \epsilon \in R^+.$ Does that help? Nov 3, 2014 at 21:33