Negation justification. $\exists x\in\Bbb R\,\forall y\in\Bbb Z^+(x<y)$

Question

$$\exists x\in\Bbb R\,\forall y\in\Bbb Z^+(x<y)$$

Which negates to

$$\forall x\in\Bbb R\,\exists y\in\Bbb Z^+(x\ge y)$$

Aren't both the statement and negation true?

Answer 1

Corrected. The original statement is true: we can take $x=0$, for instance, since $0$ is less than each positive integer. Its negation is not true: if you take $x=0$, there is no $y\in\Bbb Z^+$ such that $0\ge y$. Indeed, $\exists y\in\Bbb Z^+(x\ge y)$ is false for every $x<1$.