First prove that no positive real is less than every positive rational.

Now when 0\le

First prove that no positive real is less than every positive rational.

So when $0\leq a<b,$ take $q_0\in \mathbb Q$ with $0<q_0<\frac {b-a}{3}.$ Let $k_0$ be the least $k\in \mathbb N$ such that $kq_0>a.$ So $(k_0-1)q_0\leq a<k_0q_0.$ Therefore $$a<(k_0+1)q_0=(k_0-1)q_0+2q_0\leq a+2q_0<a+2\left(\frac {b-a}{3}\right)<b.$$