When is $bn-b\ge\lfloor bn\rfloor$?

Question

If $b\in\mathbb R\setminus\mathbb Q$ is between $0$ and $1$ then how to choose $n\in\mathbb N$ so that $$bn-b\ge\lfloor bn\rfloor>0$$

Answer 1

It is sufficient to pick a rational number $\frac mn$ slightly larger than $b$, such that $1 < m < n$ and

$$0 < \frac{m-1}{n-1} \le b < \frac mn < 1$$

Then because $\dfrac{m-1}{n} < \dfrac{m-1}{n-1}$,

$$ \frac{m-1}n < b < \frac mn\\ m-1 < bn < m\\ \lfloor bn\rfloor = m-1$$

And

$$\begin{align*} \frac{m-1}{n-1} &\le b\\ m-1 &\le bn - b \end{align*}$$

We get $$bn - b \ge \lfloor bn\rfloor$$

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One systematic way to find such $\frac mn$ is to consider the third convergents of $b$, obtained from the continued fraction of $b$:

$$\begin{align*} b &= 0 + \cfrac1{a_1 + \cfrac 1{a_2 + \cfrac 1{a_3 + \ddots}}}\\ b_3 &= \cfrac1{a_1 + \cfrac 1{a_2 + \cfrac 1{a_3}}}\\ &= \frac{a_2a_3+1}{a_1(a_2a_3+1)+a_3}= \frac mn \end{align*}$$

$b_3$ overestimates $b$, but is the closest rational approximation with $n$ or less as denominator.

In particular, $b_3$ is closer to $b$ than any rational approximation with denominator $n-1$, so considering that $\frac{m-1}{n-1} < \frac mn$,

$$\frac{m-1}{n-1} < b < \frac mn$$

Any odd convergent higher than the third also works. The first convergent does not work though because its numerator is $1$.

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The above is applicable even to some rational $b$, that have long enough continued fraction to have the third convergent ($> b$).

For general rational $b = \frac cd$, pick $b = \frac {m-1}{n-1}$ and so $n = d+1$, then

$$\begin{align*} bn &= b(d+1)\\ &= c + b\\ bn-b &= c\\ &\ge \lfloor bn\rfloor \end{align*}$$

Answer 2

If $b$ is irrational then the sequence of fractional parts of $(bn)$ is equidistributed in $[0,1)$. Choose $n$ such that the fractional part of $bn$ is in $(b,1)$. Then $bn-b>\lfloor bn\rfloor+b-b$.