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

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 2:21