Prove using the axiom of Archimedes that $\forall a, b \in \Bbb{R}, \; \exists x \in \Bbb{Q} : x \in (a, b)$
I did it in this way:
Since $x \in \Bbb{Q}$, I can write it as $\frac{l}{k}; \quad l, k \in \Bbb{Z}$
Since $\Bbb{R}$ is symmetric with respect to the origin, Is not restrictive to suppose $a, b \in \Bbb{R^+}$ and $l, k \in \Bbb{N}$.
$x \in (a, b) \implies a<x<b$
It's not difficult to prove with the axiom of Archimedes that $\exists \; x > a$ (just think $x$ as $d*f; d\in \Bbb{Q}$ and $f \in \Bbb{N}$).
Now I have to prove that $\exists \; x<b$. As I said, I can write $x$ as $\frac{l}{k}$, with $l, k \in \Bbb{N}$
$\frac{l}{k}<b \implies l< b*k$.
Since $b$ is real and $l$ and $k$ natural, the axiom of Archimedes ensures that exists an appropriate $k$ for which the inequality is verified.
I think it's right, but I'm not sure. Can you please tell me what you think?
Thanks,
Lorenzo
You've shown that, given $a,b>0$, the exists $q>a$ and $q'<b$, $q$ and $q'$ rationals.
Let $a,b$ be reals. By your arguments there exists $q$ rational such that $0<q< b-a$. By the archimedian property $nq> a$ for some natural $n$. Fix the smallest such $n$. You should be able to show that, if $a$ and $b$ are positive, then $a<nq<b$.
If $a <0$ repeat the argument for $(\max\{-b,0\}, -a)$, and argue that this is enough.