Proving $\exists k \in \mathbb{Z}: \forall l \in \mathbb{Z}: \lnot(k \mid l) $?

Question

$$\exists k \in \mathbb{Z}: \forall l \in \mathbb{Z}: \lnot(k \mid l) $$

I have no idea how to approach this problem? If true, prove it, and if false prove the the negation. With its negation being:

$$\forall k \in \mathbb{Z}:\exists l \in \mathbb{Z}: k \mid l$$

My intuition is telling me the negation is true, because for any k we can just choose k for l and then we get $k \mid k$, which is always true for the integers.

But then, I believe the orginal statement is also true. Because clearly, $ 0 \mid l $ only for $l = 0$. So, $$\forall l \in \mathbb{Z}: \lnot(0 \mid l)$$

Any suggestions of where I am going wrong?

Answer 1

According to the general definition of divisor the negation is always true assuming $l=mk$ and the original statement is not true.