# Negating statements with quantifiers

I understand that when we want to negate a statement with universal quantifier, that quantifier changes to existential quantifier, and vice versa. For example,

negation of $(\exists x\in\Bbb N)(x+1=8)$ would be $(\forall x\in\Bbb N)(x+1\neq 8)$ and the negation of $(\forall x\in\Bbb N)(x<8)$ would be ($\exists x\in\Bbb N)x\geq 8$.

However, for some reason I don't understand how negating statements with both universal and existential quantifiers works. For instance,

$(\forall x\in\Bbb R)(\exists y\in\Bbb R) x^{2}+y^{2}\geq 4$ should be negated as $(\exists x\in\Bbb R)(\forall y\in\Bbb R) x^{2}+y^{2}<4$. $(*)$

I'm not even sure what I don't understand, I know that

$\neg(\forall x \in\Bbb R)=(\exists x\in\Bbb R)$

$\neg(\exists y\in\Bbb R)=(\forall y\in\Bbb R)$

$\neg(x^{2}+y^{2}\geq 4)=x^{2}+y^{2}<4$

I guess that when reading those statements $(*)$, I don't understand how one negates the other.

• What is the question? – Bijesh K.S Oct 29 '16 at 11:41
• How would you read $(\forall x\in\Bbb R)(\exists y\in\Bbb R) x^{2}+y^{2}\geq 4$ and how would you read $(\exists x\in\Bbb R)(\forall y\in\Bbb R) x^{2}+y^{2}<4$ ? – lmc Oct 29 '16 at 11:59
• First one: For all $x\in\mathbb{R}$, there exists $y\in\mathbb{R}$ such that $x^2+y^2\geqslant 4 .$ Second one: There exists $x\in\mathbb{R}$ such that for all $y\in\mathbb{R},x^2+y^2<4.$ – Bijesh K.S Oct 29 '16 at 12:05

When you negate a quantifier, you 'bring the negation inside', e.g. $\neg \forall x P(x)$ is equivalent to $\exists x \: \neg P(x)$, where P(x) is some claim about $x$.
If you have two quantifiers, that still works the same way, e.g. $\neg \forall x \exists y P(x,y)$ is equivalent to $\exists x \neg \exists y P(x,y)$, which in turn is equivalent to $\exists x \forall y \neg P(x,y)$. And once you see that, you can understand that you can move a negation through a series of any number of quantifiers, as long as you change the quantifier: each $\forall$ becomes a $\exists$ and vice versa.