I've got to prove (or disprove) the following statement:
$\exists x \in \mathbb{N} \; \forall y \in \mathbb{N}: y \mid x$,
which translates into "It exists such $x$ from the set of natural numbers, that it is divisible by every other natural number."
What puzzles me is that (at the first sight) theoretically we can always build such $x$ by simple multiplication with $y$, but at the same time I am not really sure if we can do this in general. Let for example $y$ be bigger than $x$ (which it can be since this should work for every natural $y$); this clearly does not divide into $y$ anymore. Again by multiplication, I could find a new $x$ that would statisfy the condition... but so can I find $y$ bigger than this new $x$...
So in fact, this statement would only be true when there exists the biggest natual number that is at the same time a multiple of all the previous natural numbers. This clearly cannot be satisfied.
Could this already do as a proof that such $x$ does not exist or do I have to prove the negation of the original statement $\lnot (\exists x \in \mathbb{N} \; \forall y \in \mathbb{N}: y \mid x)$?
EDIT: Note that we distinguish between $\mathbb{N}$ and $\mathbb{N}_0$ ($0 \not\in \mathbb{N}$, but $0 \in \mathbb{N}_0$).