What are the differences between these two statements? 
*

*For every positive real number x, there is a positive real number y less
than x with the property that for all positive real numbers z, yz ≥ z.

*For every positive real number x, there is a positive real number y with
the property that if y < x, then for all positive real numbers z, yz ≥ z.
The first one is false because you can pick a counter example of x = 5 and y = 1/2, but for some reason the 2nd one is true and I'm confused as to why it is. Can't one pick the x = 5 and y = 1/2 counter example for the 2nd statement?
 A: The first is false because for $x \le 1$, there is no $y$ less than $x$ you can choose that will make the inequality work. Your counterexample is not a correct one, since for $x = 5$ there does exist such a $y$ - for example, $y = 2$.
However, the second one is true. Choose $y = x + 1$. Is $y < x$? No! So the premise of "if $y < x$, then for all positive real numbers $z$, $yz \ge z$" is never satisfied, and hence the statement as a whole is vacuously true as a result.
A: The second seems false to me.
Let $x < 1$. If there exists such $y$ and $y<x<1$. Let $z=1$; then $zy=1\times y < 1=z$. This is a contradiction!
A: If statement $P$ is known to be false the the statement $P\implies Q$ is true, because the only way for $P\implies Q$ to be false is to have the truth of $(P\land \neg Q),$  which is impossible when $P$ is false.
Consider the case where $P$ is $x>y$ and  $Q$ is "I am a unicorn." Now given any $x\in R^+,$ let $y=x+1.$ So $P$ is false. So $(P\implies Q)$ is true. A careless reading of your second proposition may give the impression that the $y$ that's asserted to exist must  satisfy $y<x. $
As far as I can tell, I am not and have never been a unicorn. 
A: 
Statement 1:
  For every positive real number $x$, there is a positive real number $y$ less than $x$ with the property that for all positive real numbers $z$, $yz \geq z$.

This statement is false.

Proof:
  Assume the statement was true. For $x < 1$ and $z = 1$ we would have
  $$
x > y = yz \geq z = 1 > x,
$$
  but $x > x$ is obviously false. So the statement cannot be true.
  $\tag*{$\square$}$

Now let us consider the second statement, which is a theorem:

Theorem 2:
  For every positive real number $x$, there is a positive real number $y$ with the property that if $y < x$, then for all positive real numbers $z$, $yz \geq z$.
Proof: Let $x > 0$. Then for every $y > 0$ with $y \geq x$, implications of the form
  $$
y < x \Longrightarrow \text{whatever statement}
$$
  are trivially true. This includes, of course, the property
  $$
y < x \Longrightarrow \forall z > 0 : yz \geq z.
\tag*{$\square$}
$$

