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:
 * $$x \,\bmod\, 1 = x - \left \lfloor {x} \right \rfloor$$