Negation of Implication

Question

For all real numbers $x$, if $x^2\ge1$ then $x > 0$.

Is the negation of the statement above is there exist real numbers x, x^2 greater than or equal to 1 and x < 0?

Answer 1

An implication $\forall(x) (P(x) \rightarrow Q(x))$ is logically equivalent to $\neg(\exists(x)(P(x) \wedge \neg Q(x)))$.

Let $P(x):=$ $x^{2} \geq 1$

Let $Q(x):=$ $x > 0$

Then the negation of $\forall(x) (P(x) \rightarrow Q(x))$ is $\exists(x)(P(x) \wedge \neg Q(x))$, which is that there exists an x such that $x^{2} \geq 1$ and $x \leq 0$

Answer 2

Your statement is

$(\forall x \in \mathbb R) \; \; ( x^2 \geq1 \implies x>0)$

$\iff$

$(\forall x \in \mathbb R) \;\;\big( (x^2<1) \lor \; \; (x>0)\big)$

so the negation is

$(\exists x\in \mathbb R) : (x^2\geq 1) \;$ and $\; (x\leq 0)$.