I have a statement that says, for all $x$ in the set of integers, there is some $y$ in the set of naturals, such that $y$ is less than $x$.
I know that this is false because $x$ can be negative, and a positive is not less than a negative.
I'm not sure how to construct the disproof for this statement.
On
A universal statement is disproved simply by a counterexample like the ones you mention. You say: take the integer -5. All natural numbers are bigger than -5, so the statement is false for -5 and thus not true for all integers.
If you think that the fact that -5 is smaller than all natural numbers needs a proof, you can do this by a trivial induction. -5 < 0 , < 0+1 ...
On
Whenever you want to disprove something do the following: take the statement, each time "for every" or something along those lines appear change it to "exists" and vice versa and change what you need to prove to it's complementary.
To disprove this you do:
For every integer $x$ exists natural $y$ such that $y<x$
Changes into:
Exists integer $x$ for every natural $y$ such that $y\ge x$
This you can prove by an example of such a number (like you said, take an example for negative $x$)
Formally it looks like:
$\lnot(\forall x\in\Bbb Z\exists y\in \Bbb N: y<x)$
This part is the original statement with not at the start, which means this statement is what you need to prove to disprove the original statement, simply it and you get:
$\lnot(\forall x\in\Bbb Z\exists y\in \Bbb N: y<x)=\exists x\in\Bbb Z\forall y\in \Bbb N: y\ge x$
$\forall=$for all
$\exists=$exists
$\in\Bbb Z=$is an integer/in the integers set
$\in\Bbb N=$is a natural/in the natural numbers set
$\lnot=$logic not
You can write it down as follows (then depending from the course you are attending it might or might not be formal enough):
Suppose $x\in\mathbb{Z}$ is negative, and let $y\in\mathbb{N}$. Then $$x<0\le y\ .$$ Therefore, there is no $y\in\mathbb{N}$ satisfying the statement.