This exercise is from Tom Apostol's Calculus Vol.1:
I 3.12 Exercises (page 29)
12) The Archimedian property of the real-number system was deduced as a consequence of the least-upper-bound axiom. Prove that the set of rational numbers satisfies the Archimedian property but not the least-upper-bound property. This shows that the Archimedian property does not imply the least-upper-bound axiom.
I don't get what I have to do, and what it means -- to satisfy the least-upper-bound-property. I know it is an axiom, I didn't know it is a property. My closest guess is that if I have two rational numbers $x < y$, I can find an integer $n$ such that $nx > y$. And I don't need to refer to that axiom, because I can find this $n$ by playing with the integers inside the rational numbers.
Previously from the book:
The Archimedian property (page 26): If $x > 0$ and if $y$ is an arbitrary real number there exists a positive integer $n$ such that $nx > y$.
The least-upper-bound axiom (page 25): Every nonempty set $S$ of real numbers which is bounded above has a supremum; that is, there is a real number $B$ such that $B = sup S$.