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 $k\in \mathbb{Z}$ ?

Hint: $a$ lies between some $\frac{1}{k}$ and $\frac{1}{k+1}$. — Dzoooks, Jul 01 '18 at 03:57
@Dzoooks When we are trying to show existence of such a $k$, how can you claim such a thing. — Our, Jul 01 '18 at 04:02
Note $(0,1] = \sqcup_{k \geq 1} \left(\frac{1}{k+1},\frac{1}{k}\right],$ and $a \in (0,1]$. — Dzoooks, Jul 01 '18 at 04:12

score 1 · Accepted Answer · answered Jul 01 '18 at 04:06

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 $1/k$.

Peter Szilas · Answer 2 · 2018-07-01T14:10:16.283

0

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 $m \ge 1$..

edited Jul 01 '18 at 14:10

answered Jul 01 '18 at 14:05

Peter Szilas

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

2 Answers2