The statement is :
$\forall x \in \mathbb{Z}$, $\exists y \in \mathbb{R}, (x\geq y \Rightarrow \frac{x}{y}=1)$.
I think the negation without negation symbol will be as follows :
$\exists x \in \mathbb{Z}$, $\forall y \in \mathbb{R}, (x\geq y \Rightarrow \frac{x}{y} \neq 1)$.
Is it correct.
And also the statement seems false as $x=0$ is counterexample.
Am I correct?
Best Answer
No, it is not. The negation of $A\implies B$ is $A\wedge\neg B$. So, the negation of your statement is$$\exists x\in\Bbb Z,\forall y\in\Bbb R,\left(x\geqslant y\wedge\frac xy\ne1\right).$$And the original statement is true: if $x\in\Bbb Z$, take $y>\max\{x,0\}$. Then both $x\geqslant y$ and $\frac xy=1$ are false, and therefore $x\geqslant y\implies\frac xy=1$.