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

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?

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 16 '18 at 23:28
  • $\begingroup$ Oh so your saying that since $\exists l\in \mathbb{Z}: 0 \mid l.$ We cannot conclude $\forall l \in Z: \lnot (0 \mid l)$ $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$ Nov 16 '18 at 23:50
  • $\begingroup$ @FefnirWilhelm Yes you are right, with reference to the general definition the negation is true. $\endgroup$
    – user
    Nov 17 '18 at 8:15

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