*NOT* [quantifier]. Why is *NOT* [for every] equivalent to [there is]? Given a proposition such as, "For every real number $x \ge 2$, $x^2 + x - 6 \ge 0$", I am told that the negation, "NOT [For every real number $x \ge 2$, $x^2 + x - 6 \ge 0$]", would be "There is a real number $x \ge 2$ such that $x^2 + x - 6 < 0$".
I am specifically confused with regards to why NOT [for every] is equivalent to [there is] rather than [for none]? It seems logical to me that the negation of everything (for all) should actually be equivalent to nothing?
I would greatly appreciate it if someone could please take the time to clarify this concept.
 A: You do not have equivalence between 'not for every' and 'there is'. You also need to negate the open sentence which is in the scope of the quantifiers. And you in fact did that: you turned the inequality around.
The general rule is that $\neg\forall xP(x)$ is equivalent to $\exists x\neg P(x)$. And that is quite intuitive: that not everything is a $P$-thing is the same as saying that there is a non-$P$-thing.
A little more detail: Your statement has more logical structure then that, as it contains a conditional. The statement has the form $\neg\forall x(Q(x)\to R(x))$ with $Q(x)$ being $x\ge2$ and $R(x)$ being $x^2 + x - 6 \ge 0$. This is equivalent to $\exists x\neg(Q(x)\to R(x))$ which is again equivalent to $\exists x(Q(x)\wedge\neg R(x))$. And $\neg R(x)$ is equivalent to $x^2 + x - 6 < 0$.
A: As an addendum to Casper's answer: there are logics in which $\neg (\forall x) P(x)$ is not equivalent to $(\exists x) [\neg P(x)]$.

For example (and you might like to look up "constructive mathematics" for more on this), take $(\exists x)$ to mean "there is $x$ and moreover we can in principle compute an explicit such $x$", rather than the usual "there is $x$". Let $P(x)$ (where $x$ is restricted to be an integer between $0$ and $9$) be the statement that the digit $x$ appears only finitely often in the decimal expansion of $\pi$.
Then the statement $\neg (\forall x) P(x)$ is certainly true - if not, $\pi$ would have a finite decimal expansion. But $(\exists x) [\neg P(x)]$ is not currently known to be true, because we don't know of any digit which definitely appears infinitely often in $\pi$'s decimal expansion.
