Archimedean Principle
Andrew Mclaughlin 
Prove the Archimedean principle for $\mathbb{Q}$, the set of rational numbers. I know the proof for real numbers If not, then the nonempty set $S = \{nx \mid n ∈ \mathbb{N}\}$ has an upper bound $y$, and so by the least upper bound property of the reals, it has a least upper bound $u_∗$. But then $u_∗ − x < u_∗$ (since $x$ is positive) and so it follows that $u_∗ − x$ cannot be an upper bound of $S$. Hence there exists a natural number $m$ such that $u_∗ − x < mx$, that is, $u_∗ < (m + 1)x \in S$. This contradicts the fact that $u_∗$ is an upper bound of $S$.
$\endgroup$ 12 Answers
$\begingroup$Proving the Archimedean principle first for ${\mathbb R}$, using the $\sup$, is in a way cheating. This principle is already present in ${\mathbb N}$ and should be proven from the Peano axioms.
Archimedean principle for ${\mathbb N}$: Given natural numbers $x\geq 1$ and $y\geq0$ there is an $n\geq1 $ such that $n\>x>y$.
Proof. This is true for $y=0$ and all $x\geq1$. Assume that it is true for some $y\geq0$ and all $x\geq1$. Consider now $y+1$ and an arbitrary $x\geq1$. By the induction assumption there is an $n\geq1$ with $n\>x>y$, and putting $n':=n+1$ we have $$n'\>x=n x+x>y+x\geq y+1\ .$$
It is not difficult to extend this from ${\mathbb N}$ to ${\mathbb Q}_{>0}$.
$\endgroup$ 1 $\begingroup$The Archimedean property for $\mathbb Q$ says that :
for all $r_1,r_2 \in \mathbb Q$, there exists $n \in \mathbb N$ such that : $nr_1 \ge r_2$.
We may prove it by contradiction, assuming that :
there exists $r_1,r_2 \in \mathbb Q$ such that, for all $n \in \mathbb N$ : $nr_1 < r_2$.
We have $p_1,p_2,q_1,q_2 \in \mathbb Z$ such that : $r_1=\frac {p_1} {q_1}$ and $r_2=\frac {p_2} {q_2}$, and we can assume that $q_1,q_2 \in \mathbb Z^+$.
Then the above inequality gives us :
$n(q_2p_1) < (q_1p_2)$, for all $n$.
Thus, we have found $M_1,M_2 \in \mathbb N$ such that : $nM_1 < M_2$.
By uniqueness of quotient and reminder of the division in $\mathbb N$ we have that :
$M_2 = M_1k+h$, with $h < M_1$.
Thus :
$nM_1 < M_1k+h < M_1k+M_1 = M_1(k+1)$, for all $n$.
Now we let $n=(k+1)$ and we finally reach a contradiction :
$\endgroup$$M_1(k+1) < M_1(k+1)$.
