Floor function inequality: $\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor<1$

Question

I would like to dissect the following inequality to figure out its properties.

$$\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor>1$$

where $n>0$ and $0<x<n^2$ and $x$ is an integer.

I would like to know will $\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor<1$ for all $x>2n+1$?

It is often useful to write $\lfloor x \rfloor + {x} =x$, the sawtooth function ${x}$ is bounded and thus easier to handle — vonbrand, May 14 '14 at 17:15
The $> 1$ inequality does not hold when $x$ divides $2n+1$ or $n^2$. — NovaDenizen, May 15 '14 at 02:21

score 2 · Answer 1 · answered Aug 30 '22 at 20:19

(Getting this off the unanswered list.)

For any $r\in\Bbb R$, $r-\lfloor r\rfloor$ is the fractional part of $r$, usually written $\{r\}$. Thus, you’re asking about

$$\left\{\frac{n^2}x\right\}-\left\{\frac{2n+1}x\right\}\,.$$

Clearly

$$0\le\left\{\frac{n^2}x\right\},\left\{\frac{2n+1}x\right\}\le\frac{x-1}x=1-\frac1x<1$$

for any positive integer $x$, so

$$\frac1x-1\le\left\{\frac{n^2}x\right\}-\left\{\frac{2n+1}x\right\}\le 1-\frac1x<1$$

for any positive integer $x$.

Floor function inequality: $\frac{n^2}{x}-\left\lfloor\frac{n^2}{x}\right\rfloor+\frac{2n+1}{x}-\left\lfloor\frac{2n+1}{x}\right\rfloor<1$

1 Answers1