How to prove that for $0 < a \leq 1$ and $a \not = 1/n$ there exists the maximal $k$ s.t $k\cdot a < 1$

Question

In this article of IIME Journal, at page 16, it is claimed that

for $0 < a \leq 1$ and $a \not = 1/n$ for any integer $n$, there exists an integer $k$ such that $k\cdot a < 1$ and $k$ is the greatest integer satisfying this condition.

However, how can we show the existence of such a $k\in \mathbb{Z}$ ?

Answer 1

Since $a>0$, there is some positive integer $N$ such that $a>1/N$. Then for any $m\geq N$, $ma>1$. So, there are only finitely many positive integers $k$ such that $ka\leq 1$ (and there exists at least one such $k$, namely $k=1$). Any finite nonempty subset of a totally ordered set has a greatest element, so there is a greatest positive integer $k$ such that $ka\leq 1$. Finally, we cannot have $ka=1$ for this $k$ since then $a$ would be equal to $1/k$.

Answer 2

Correct me if wrong:

Rephrasing :

$0<a<1$ , since $a= 1 =1/1$ is excluded.

$S:=${$k | k \in \mathbb{Z}$ s.t. $ka <1$}, or

$S= ${$k|k \in \mathbb{Z}$ s.t. $k<1/a$}.

Note: $S \not = \emptyset$ since $k=1 \in S$.

$S$ is bounded above by $1/a$, $\sup(S)$

exists, which implies that $S$ has a

maximal element $m \ge 1$..