# Analysis question about density of rationals

I am looking through a proof of the density property.

We are given that

$x < y$

by the Archimedean property, there exists a positive integer n such that;

$n(y-x) > 1$ or $\frac{1}{n} <(y-x)$

There exists an integer $m$ such that

$m \leq nx < m+1 \implies \frac{m}{n} \leq x < \frac{m+1}{n} < y$

The last part, i don't see how one can say

$\frac{m+1}{n} < y$

What i can understand is that the distance $d(y,x) > 1/n$ but from that how do you deduce that $\frac{m+1}{n} < y$

• d(y,x) is actually greater than 1/n, which is what allows you to deduce that (m+1)/n is less than y. – Jesse Feb 17 '15 at 0:03

Note that from $n(y-x)>1$, we have $\color{green}{x+1/n}\color{red}{<y}$.

Since $m\le nx<m+1$, we have $m/n\le\color{blue}{x<(m+1)/n}$. In particular, we have $m/n\le x$; so adding $1/n$ we obtain $m/n+1/n=\color{blue}{(m+1)/n\le}\color{green}{x+1/n}\color{red}{<y}$. Thus we conclude $x<(m+1)/n<y$, as desired.

By other hand, you can see geometrically.

$$|-\;\;\color{red}{1/n}\;\;------\;\;\color{blue}{m/n}\;\;--\color{green}{|----|}$$ $$0\phantom{--------------}\;\;\;\;x\phantom{----}y$$

By archimedean principle, $\color{red}{1/n}<\color{green}{y-x}$.

By well-ordeing principle there exists $k=\min\{j\in\mathbf N:j/n\ge x\}$. We define $m:=k-1$, so clearly we have $\color{blue}{m/n}<x$.

Adding both $\color{red}{1/n}<\color{green}{y-x}$ and $\color{blue}{m/n}<x$, we obtain $1/n+m/n<(y-x)+x$, i.e., $(m+1)/n<y$.

• This is exactly what i was after. Thanks ever so much. – user214138 Feb 17 '15 at 9:02
• You welcome! :-) – Cristhian Gz Feb 17 '15 at 14:18