Question

(Density property) if a, b ∈ R with a < b, then there exists a rational number r ∈ Q such that a < r < b.

Answer 1

Here as $b-a>0$ , therefore using the Archimedean property we have, $n\in \Z_{>0}$ such that

$n(b-a)>1$ .

Now, let $m=[na]$ (g.i.f). Then $m\leq na<m+1$ . Also we have $m+1<ny$ . (To get this you can assume $m+1\geq ny$ and arrive at a contradiction)

Thus we have,

$$m\leq na<m+1<nb$$

Dividing all sides by $n$,

$$\frac{m}{n}\leq a < \frac{m+1}{n}<b$$

Therefore as $m,n\in\Z$ , we have $r=\frac{m+1}{n}\in \mathbb{Q}$ as the rational number between $a$ and $b$ .