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 $x$.

Also see

• Definition:Fractional Part
• Definition:Modulo 1: $x \bmod 1 = x - \floor x$

Sources

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