# *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.

• Not always true = sometimes false. – Michael Burr Feb 6 '17 at 4:14
• If it's not the case that I wake up early every day, does that mean thay I never wake up early? – YoTengoUnLCD Feb 6 '17 at 4:22
• @YoTengoUnLCD Your analogy makes sense. – The Pointer Feb 6 '17 at 4:23

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 $x^2 + x - 6 < 0$.
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.