Archimedean property of the rational numbers Can you prove the Archimedean property of the rational numbers without constructing the reals and using the least upper bound property? It seems odd to have to take this roundabout approach, but I don't know any proof that avoids it.
 A: Let $x \in \mathbb{Q}$ 
If $x < 0$ there's nothing to do.
If $x \geqslant 0$ then we can write $x$ as $\dfrac{n}{m}$ (with no common factor other than 1), such that $n \in \mathbb{N}$ and $m \in N_{\ne 0}$
Since $m \geqslant 1$ then we have $n = mx \geqslant x$ which means $n + 1 > x$
A: I came across this question in Mr. Apostol's Calculus.
My answer is a bit weird but it seems to work.
Assume the set of natural numbers is up bounded by a certain rational number a. It's certain that a >= 1 because 1 is the first member of natural numbers. Then a can be represented by two natural numbers p and q in the form of p/q.  
(1) if q = 1, then a = p. Hence p + 1, which has to be natural number, is above the upper bound.
(2) if q >1, then a < p. hence p, a natural number from the assumption is above the upper bound.
On the other hand, it's easy to find a set of rational numbers bounded above and does not have a least upper bound.
Thus Archimedean property does not imply the completeness axiom, a.k.a, the least-upper-bound axiom
QED.
A: Every positive rational number is of the form $m/n$ where $m$, $n$ are positive integers.  If you add up more than $n$ copies of this, the sum is more than 1, so there you have the Archimedean property.
A: Let $x$ be a positive rational number. Then we can write $x=\frac{m}{n}$ with $m,n\in\mathbb{N}$ and $n\ne0$. Furthermore, if $x>0$ then $m\ge1$.
Multiplying $m$ by some positive integer $p$, we get $mp\ge p$. Certainly we can say $mp>\frac{p}{q}$ for some positive integer q. Let $y=\frac{p}{q}$.
Since $x=\frac{m}{n}$, $nx=m$. Hence, $npx=mp$. Let $np=z\in\mathbb{N}$. Then for $x,y\in\mathbb{Q}$, $x>0$, $\exists z\in\mathbb{N}$ s.t. $zx>y$, and we are done.
But what if $q=1$ and $y=p$? Then we have $m\ge1\rightarrow mp+m>y\rightarrow m(p+1)>y\rightarrow nx(p+1)>y\rightarrow (np+n)x>y$ and $(np+n)$ is the positive integer we are looking for.
