density of the rationals on the reals proof The Archimedean property,
If $x\in R, y\in R, x>0,$ then $\exists n \in Z^+ $ such that $nx>y$
Now we want to prove that the rationals $Q$ are dense on the reals $R$.
The proof in the book Introduction to Real Analysis by Bartle and Sherbet start their proof by assuming that $x>0$. Why do they do this? In fact the Archimedean property like this where x  needs be greater than 0 is strange.
In other text books they say the AP is
if $x\in R, \implies n_x \in N $ st $x \leq n_x$
which makes more sense in the context.
Basically, why do they assume that $x >0$?
 A: The meaning of $x$ is different between the two versions of the Archimedean property you're quoting. I will switch to different notation so that you can see the similarity.
The second property says:

For all $r \in \mathbb R$, there is an $n \in \mathbb N$ such that $n > r$. Every real number has an integer bigger than it.

The first property says:

For all $r \in \mathbb R$, and for all $u \in \mathbb R$ with $u > 0$, there is an $n \in \mathbb N$ such that $nu > r$. Every real number has an integer multiple of $u$ bigger than it.

The second property is a special case of the first, with $u=1$. (Of course, you can also use the second property to prove the first, by applying it to $\frac ru$.)
The intuition behind the Archimedean property is this: all real numbers live in the same world. If you pick any real number as your unit, you will be able to measure all other real numbers in those units.
But we want our unit $u$ to be a positive real number, because we want $nu$ to get arbitrarily large as $n \in \mathbb N$ increases. This would not be necessary if we took $n \in \mathbb Z$ instead.
