I am having a hard time coming to grips with one assumption of this proof
Let $x$ and $y$ $\in\mathbb{R+}$ and without loss of generality x<y.
We pick a natural number large enough to make $\frac{1}{n} < y-x$ ,
for $n\in\mathbb{N}$ (Archimedes axiom).Now let's call $k$ the maximum natural number such that $\frac{k}{n}<= x$, then
$x<\frac{k+1}{n}$ (well-ordering-principle).At this point we can also know $\frac{k+1}{n}<y$ (you can prove it by contradiction).
Thus $z = \frac{k+1}{2}$ is a rational number that satisfies $x<z<y$ where $x$ and $y$ $\in\mathbb{R}$
My question is the assumption of $k$ such that $\frac{k}{n}<=x$. Why is this a valid assumption? It seems a little too much to use $n$ as the denominator, what would be a more meticulous approach (if this is not rigorous enough)?
Best Answer
If you want to have a rigorous proof that works for arbitary real numbers :
WLOG, we have $\ x<y\ $.
Define $\ \epsilon:=y-x\ $ which is clearly positive and let $n$ be an integer with $\ n>\frac{2}{\epsilon}\ $
Then, we have $\ ny-nx=n(y-x)=n\epsilon>2\ $ , hence there is an integer $\ m\ $ with $\ nx<m<ny\ $. So we have $\ x<\frac{m}{n}<y\ $ as desired.