Sum of Floors not greater than Floor of Sum

Theorem
Let $\left \lfloor {x} \right \rfloor$ be the floor function.

Then:
 * $\left \lfloor {x} \right \rfloor + \left \lfloor {y} \right \rfloor \le \left \lfloor {x + y} \right \rfloor$

The equality holds:
 * $\left \lfloor {x} \right \rfloor + \left \lfloor {y} \right \rfloor = \left \lfloor {x + y} \right \rfloor$

iff:
 * $x \,\bmod\, 1 + y \,\bmod\, 1 < 1$

where $x \,\bmod\, 1$ denotes the modulo operation.

Proof
From the definition of the modulo operation, we have that:
 * $x = \left \lfloor {x}\right \rfloor + \left({x \, \bmod \, 1}\right)$

from which we obtain:

Hence the inequality.

The equality holds iff:
 * $\left \lfloor {\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right)} \right \rfloor = 0$

that is, iff:
 * $x \,\bmod\, 1 + y \,\bmod\, 1 < 1$.