Proof of a lemma.

Question

In my textbook, the author offers a proof for the density of the rationals within the reals (where the reals are a ordered Archmediean field with completeness), and proves that they're everywhere dense by saying that for $0 \leq a \leq b$ (for real $a$ & $b$) there is a positive integer $n$ such that ${1\over n } \leq b-a$ and also that $a < m({1 \over n})$ for a positive integer $m$. Then, the author says that we assume that there exists a $k \over n$ ($k \in \mathbb{Z}_+$) which is the least that can satisfy $a < m({1 \over n})$. The rest is irrelevant to my question but the proof hinges on this "least ${k \over n}$" where by $a < {k \over n}$ and ${{k-1} \over n} \leq a < {k \over n}$, but I've no idea if it exists or not (in terms of proof). Does someone know how to prove it?

Answer 1

I think it can be restated as: For each real number $b$ and each natural number $n$, there is a unique integer $m$ such that $m/n$ is in the interval $[b – 1/n, b)$.

For such an $m/n$, we would have $b – 1/n \leq m/n < b$, or $nb - 1 ≤ m < nb$.