# Real Number minus Floor

## Theorem

Let $x \in \R$ be any real number.

Then:

$x - \floor x \in \hointr 0 1$

where $\floor x$ is the floor of $x$.

That is:

$0 \le x - \floor x < 1$

## Proof

 $\ds \floor x$ $\le$ $\, \ds x \,$ $\, \ds < \,$ $\ds \floor x + 1$ Definition of Floor Function $\ds \leadsto \ \$ $\ds \floor x - \floor x$ $\le$ $\, \ds x - \floor x \,$ $\, \ds < \,$ $\ds \floor x + 1 - \floor x$ subtracting $\floor x$ from all parts $\ds \leadsto \ \$ $\ds 0$ $\le$ $\, \ds x - \floor x \,$ $\, \ds < \,$ $\ds 1$ $\ds \leadsto \ \$ $\ds x - \floor x$ $\in$ $\, \ds \hointr 0 1 \,$ $\ds$ as required

$\blacksquare$

## Also denoted as

The expression $x - \floor x$ is sometimes denoted $\fractpart x$ and called the fractional part of $x$.