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:

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

$\ds \floor x$

$\le$

$\, \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

$\blacksquare$

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

1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10.4 \ \text{(ii)}$: The well-ordering principle