How to find floor $\frac{(x+\frac{1}{2})} {x}$?

Question

Not sure how to figure out this...I tried just doing normal $\frac{(x+\frac{1}{2})} {x}$ but I know that's not solving for the floor and I know I need a squeeze theorem..but not sure how to solve this one..thanks.

Answer 1

$$ \left \lfloor \frac {x+\frac 12}x \right \rfloor = \left \lfloor 1 + \frac 1{2x} \right \rfloor = 1 + \left \lfloor \frac 1{2x} \right \rfloor = \left \{ \begin{array}{lcl} 1 & \text{if} & x \in \left (+\frac 12, +\infty \right ) \\ 0 & \text{if} & x \in \left (-\infty, -\frac 12 \right ] \\ m & \text{if} & m-1 \le \frac 1{2x} < m \\ -m & \text{if} & -m-1 \le \frac 1{2x} <-m \end{array}\right . $$ where $m \ge 1, m \in \mathbb N$

Answer 2

As $x-1\leq \operatorname{floor}(x+\frac{1}{2})\leq x+1 $ So you have $$ \frac{x-1}{x}=1-\frac{1}{x}\leq \frac{\operatorname{floor}(x+\frac{1}{2})}{x} \leq \frac{x+1}{x}=1+\frac{1}{x}$$

Answer 3

$$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=\left\lfloor1+\frac1{2x}\right\rfloor=1+\left\lfloor\frac1{2x}\right\rfloor.$$

Note that $n=\left\lfloor\frac1{2x}\right\rfloor$ if and only if $n\le\frac1{2x}<n+1.$ Hence, $0\leq \frac1{2x}<1$ if and only if $1<2x$ if and only if $x>\frac12$, in which case $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=1+0=1.$$ Also, $-1\le\frac1{2x}<0$ if and only if $2x\le -1$ if and only if $x\le-\frac12$, in which case $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=1+-1=0,$$ so the end behavior is easy to calculate. If you want more detail, then for integers $n\neq0,-1$, we have $$n\leq\frac1{2x}<n+1$$ if and only if $$\frac1{n+1}<2x\le\frac1n$$ if and only if $$x\in\left(\frac1{2(n+1)},\frac1{2n}\right].$$ Hence, $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=\begin{cases}1 & x\in\left(\frac12,\infty\right),\\0 & x\in\left(-\infty,-\frac12\right],\\1+n & n\in\Bbb Z\smallsetminus\{-1,0\},x\in\left(\frac1{2(n+1)},\frac1{2n}\right].\end{cases}$$