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

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

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?

• You go wrong after "But then". You first say that $0 | 0$, but then say that $\forall l \in \mathbb{Z}: \lnot(0|l)$. These statements are not compatible. Nov 16 '18 at 23:28
• Oh so your saying that since $\exists l\in \mathbb{Z}: 0 \mid l.$ We cannot conclude $\forall l \in Z: \lnot (0 \mid l)$ Nov 16 '18 at 23:31

According to the general definition of divisor the negation is always true assuming $$l=mk$$ and the original statement is not true.
• If we let the definition of "k divides l" be: $$\exists n \in \mathbb{Z} : l = k \times n$$ then the negation statement works for $0$, no? Nov 16 '18 at 23:50