Real Number minus Floor

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

Then:
 * $x - \left \lfloor {x} \right \rfloor \in \left[{0 . . 1}\right)$

where $\left \lfloor {x} \right \rfloor$ is the floor function of $x$.

That is:
 * $0 \le x-\left \lfloor {x} \right \rfloor < 1$

Proof
$\left \lfloor {x} \right \rfloor \le x < \left \lfloor {x} \right \rfloor + 1$ from Range of Values of Floor Function.

$\left \lfloor {x} \right \rfloor - \left \lfloor {x} \right \rfloor \le x - \left \lfloor {x} \right \rfloor < \left \lfloor {x} \right \rfloor + 1 - \left \lfloor {x} \right \rfloor$ by subtracting $\left \lfloor {x} \right \rfloor$ from all parts.

$0 \le x - \left \lfloor {x} \right \rfloor < 1$

So $x - \left \lfloor {x} \right \rfloor \in \left[{0. . 1}\right)$ as desired.

Notation
The expression $x - \left \lfloor {x} \right \rfloor$ is sometimes denoted $\left\{{x}\right\}$ and called the fractional part of $x$.

Also see the definition of modulo 1:
 * $x \bmod 1 = x - \left \lfloor {x} \right \rfloor$