Real Number minus Floor

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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

\(\displaystyle \floor x\) \(\le\) \(\, \displaystyle x \, \) \(\, \displaystyle <\, \) \(\displaystyle \floor x + 1\) Definition of Floor Function
\(\displaystyle \implies \ \ \) \(\displaystyle \floor x - \floor x\) \(\le\) \(\, \displaystyle x - \floor x \, \) \(\, \displaystyle <\, \) \(\displaystyle \floor x + 1 - \floor x\) subtracting $\floor x$ from all parts
\(\displaystyle \implies \ \ \) \(\displaystyle 0\) \(\le\) \(\, \displaystyle x - \floor x \, \) \(\, \displaystyle <\, \) \(\displaystyle 1\)
\(\displaystyle \implies \ \ \) \(\displaystyle x - \floor x\) \(\in\) \(\, \displaystyle \hointr 0 1 \, \) \(\displaystyle \) as required

$\blacksquare$


Also denoted as

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


Also see


Sources