MathDB

Find the maximum number

c

such that for all

n \in \mathbb{N}

to have

\{n \cdot \sqrt{2}\} \geq \frac{c}{n}

where \{n \cdot \sqrt{2}\} \equal{} n \cdot \sqrt{2} \minus{} [n \cdot \sqrt{2}] and

[x]

is the integer part of

x.

Determine for this number

c,

all

n \in \mathbb{N}

for which \{n \cdot \sqrt{2}\} \equal{} \frac{c}{n}.

Problem Statement