Any ideas on this problem? Thanks!
Let a be a positive number. Prove that for each real number x there is an integer n such that na is less than or equal to x is less than (n+1)a.
You know that there is an integer $\displaystyle n$ such that $\displaystyle na>x$ - by the Archimedean property. Consider the set $\displaystyle S=\{ n | na \leq x \}$. This set must be bounded by what we have just set above. Let $\displaystyle N = \max\{ S\}$. Then we have that $\displaystyle Na \leq x$, but by construction $\displaystyle N+1\not \in S$ since it is larger than the maximum and so $\displaystyle (N+1)a > x$.